Math II · A-SSE.3.b

Completing the Square to Reveal Maximum and Minimum Values

This objective teaches students how to rewrite a quadratic so its turning point is visible. That turning point is the answer to many real questions: the greatest height, the largest profit, the smallest cost, the shortest distance, the safest setting, or the most efficient design.

Concept Algebra
Domain Seeing Structure in Expressions
Read time 10 minutes

What this learning objective is really asking you to learn

A quadratic expression can be written in more than one useful form. Standard form, \(ax^2 + bx + c\), is often convenient for calculation because it displays the coefficients of the squared term, linear term, and constant term. Factored form, such as \(a(x - r)(x - s)\), is powerful because it shows the zeros. Vertex form, \(a(x - h)^2 + k\), is powerful because it shows the turning point. Completing the square is the algebraic method that turns standard form into vertex form.

The phrase “complete the square” sounds like an old classroom slogan, but it is very literal. The expression \(x^2 + bx\) is almost a perfect square. A perfect square has the form \((x + m)^2\), which expands to \(x^2 + 2mx + m^2\). To make \(x^2 + bx\) into a perfect square, we need \(2m = b\), so \(m = b/2\). The missing constant is therefore \((b/2)^2\). When students add and subtract that amount, they are not changing the value of the expression; they are reorganizing it so that a square becomes visible.

For example, start with \(x^2 + 8x + 3\). The expression \(x^2 + 8x\) would become a perfect square if we added 16, because half of 8 is 4, and \(4^2 = 16\). So we rewrite:

\[x^2 + 8x + 3 = (x^2 + 8x + 16) - 16 + 3 = (x + 4)^2 - 13\].

The expression is now in vertex form. If this is the function \(f(x) = (x + 4)^2 - 13\), the smallest possible value of \((x + 4)^2\) is 0, because any real square is never negative. That happens when \(x = -4\). Therefore the minimum value of the function is -13, and the vertex is \((-4, -13)\). We did not need a graphing calculator to find it. We did not need to test many values. The structure told us.

This is the core of the objective: produce an equivalent form that reveals a property of the quantity being represented. The algebra is not done for decoration. The new form answers a question. In standard form, \(x^2 + 8x + 3\) hides the minimum. In vertex form, \((x + 4)^2 - 13\) announces it.

The same idea works when the leading coefficient is not 1, but students must be more careful. For \(2x^2 - 12x + 5\), first factor the 2 out of the quadratic and linear terms: \(2(x^2 - 6x) + 5\). Half of -6 is -3, and \((-3)^2 = 9\), so \(x^2 - 6x\) becomes \((x - 3)^2 - 9\). Then:

\[2(x^2 - 6x) + 5 = 2[(x - 3)^2 - 9] + 5 = 2(x - 3)^2 - 18 + 5 = 2(x - 3)^2 - 13\].

Now the minimum is clear. Since \(2(x - 3)^2\) is never negative, the smallest value occurs when \(x = 3\), and the minimum output is -13. If the leading coefficient were negative, the square term would be multiplied by a negative number, and the vertex would be a maximum instead of a minimum.

Why students should learn this math

Students often ask why they need another way to rewrite a quadratic. They already learned factoring. They may have learned the quadratic formula. Why complete the square? The answer is that different forms reveal different truths. Factoring reveals where a quadratic equals zero. The quadratic formula finds solutions even when factoring is hard. Completing the square reveals the center of the curve: the vertex.

That center matters because many real problems are not asking “When does this equal zero?” They are asking “When is this as high as possible?” or “When is this as low as possible?” In real life, these are optimization questions. A business wants to maximize profit. An engineer wants to minimize material waste. A coach wants to find the launch angle or position that produces the best result. A city planner wants to minimize travel distance or cost. A student wants to choose the most efficient study plan. Whenever a quantity increases, slows, turns, and then decreases, a quadratic may be a reasonable model, and the vertex becomes the most important feature.

Suppose a small theater models its daily profit with \(P(t) = -40(t - 6)^2 + 1440\), where \(t\) is the ticket price above some base price. The expression tells us immediately that the largest profit is 1440, because \(-40(t - 6)^2\) is never positive. The maximum occurs at \(t = 6\). The square term measures the penalty for moving away from the best price. The coefficient -40 tells how quickly profit falls as the price moves away from the optimum. A student who understands vertex form can read the business story directly from the formula.

In projectile motion, a simplified height model might look like \(h(t) = -16t^2 + 64t + 5\), where \(t\) is time in seconds and \(h\) is height in feet. Completing the square gives \(h(t) = -16(t - 2)^2 + 69\). That form says the projectile reaches a maximum height of 69 feet at 2 seconds. This is not a fake school problem; it is exactly the kind of reasoning used in physics, sports analysis, safety design, and animation. A quadratic model is a machine for turning motion into a curve, and completing the square opens the machine so the peak is visible.

The bigger reason is even more important: students are learning that mathematics is not just about getting answers. Mathematics is about changing representation until the relevant truth becomes visible. This habit is valuable far beyond quadratics. In advanced algebra, students rewrite rational expressions to reveal asymptotes. In trigonometry, they rewrite identities to reveal periodic structure. In statistics, they transform data to reveal patterns. In calculus, they rewrite functions to understand rates and extremes. Completing the square is one of the first major places where students see that form is meaning.

The historical machinery behind completing the square

Completing the square is ancient. Long before students saw symbolic notation like \(x^2 + bx + c\), mathematicians solved problems involving squares, rectangles, and unknown lengths. Babylonian mathematicians, working more than three thousand years ago, used procedures that are recognizably related to completing the square. They often described quadratic problems geometrically: a square area plus a rectangular strip, or a total area made from parts. To solve the problem, they would effectively add a small square to make a larger complete square, then take a square root.

This geometric origin is worth making visible to students. Imagine a square with side length \(x\), so its area is \(x^2\). Add a rectangular strip of area \(bx\). Split that strip into two equal strips, each with dimensions \(x\) by \(b/2\), and place one strip on the right side of the square and one strip on the bottom. The shape is almost a larger square. The only missing piece is a corner square with side length \(b/2\), whose area is \((b/2)^2\). Adding that corner completes the square. The algebraic rule is not arbitrary; it is a compressed version of a geometric construction.

Greek mathematics explored geometric relationships with enormous rigor, while later Islamic mathematicians developed systematic algebraic methods. Al-Khwarizmi, whose name gives us the word “algorithm,” treated quadratic equations through geometric completion. The modern symbolic version emerged gradually as algebraic notation improved. Once expressions could be written compactly, completing the square became a portable technique for solving quadratics, analyzing curves, and deriving formulas.

The method became even more powerful after analytic geometry connected equations to graphs. A quadratic expression was no longer just an algebra problem; it described a parabola on a coordinate plane. Completing the square moved the parabola into a centered form. In coordinate geometry, the same method is used to identify circles, ellipses, and hyperbolas from general quadratic equations. For example, \(x^2 + y^2 - 6x + 4y - 12 = 0\) can be completed in both \(x\) and \(y\) to reveal a circle’s center and radius. That is the same machinery students are learning here, extended into two dimensions.

Historically, completing the square is a bridge between geometry, algebra, and functions. It begins with cutting and rearranging shapes. It becomes a symbolic method for solving equations. It becomes a way to read graphs. It becomes a tool for optimization and conic sections. This is exactly the kind of idea students need to see as part of the big map: the same machine reappears in different rooms of mathematics.

Technical execution: how the method works

The simplest case begins with \(x^2 + bx + c\). The steps are:

  1. Take half of \(b\).
  2. Square it.
  3. Add and subtract that square inside the expression.
  4. Rewrite the trinomial as a squared binomial.
  5. Combine constants.

For \(x^2 - 10x + 7\), half of -10 is -5, and the square is 25. Then:

\[x^2 - 10x + 7 = (x^2 - 10x + 25) - 25 + 7 = (x - 5)^2 - 18\].

The vertex is \((5, -18)\), and because the leading coefficient is positive, the function has a minimum value of -18.

For \(ax^2 + bx + c\), where \(a\) is not 1, factor \(a\) out of the first two terms before completing the square. For \(-3x^2 + 18x + 4\), write:

\[-3(x^2 - 6x) + 4\].

Half of -6 is -3, and the square is 9. So:

\[-3[(x^2 - 6x + 9) - 9] + 4 = -3(x - 3)^2 + 27 + 4 = -3(x - 3)^2 + 31\].

The vertex is \((3, 31)\). Because the coefficient of the square term is negative, the function has a maximum value of 31.

Students should understand why the sign of \(a\) controls maximum or minimum. The expression \((x - h)^2\) is always greater than or equal to zero. If \(a > 0\), then \(a(x - h)^2\) is also greater than or equal to zero, so the smallest output is \(k\). The parabola opens upward. If \(a < 0\), then \(a(x - h)^2\) is less than or equal to zero, so the largest output is \(k\). The parabola opens downward. The vertex \((h, k)\) is the point where the squared distance from \(h\) is zero.

Completing the square also explains the axis of symmetry. In \(a(x - h)^2 + k\), inputs equally far from \(h\) give the same output because the square removes the direction. For example, if \(h = 3\), then \(x = 2\) and \(x = 4\) are both one unit from 3, so \((x - 3)^2\) is 1 for both. This is why the vertical line \(x = h\) is the axis of symmetry.

The method should not be taught as disconnected from graphing. Each algebraic move has a visual meaning. The expression \(a(x - h)^2 + k\) describes a basic parabola \(y = x^2\) that has been shifted horizontally by \(h\), shifted vertically by \(k\), stretched or compressed by \(a\), and possibly reflected if \(a\) is negative. Completing the square turns a hidden transformation into a visible one.

What this math represents in real life

Completing the square represents the search for a best point in a curved relationship. In a quadratic model, the best point is often the vertex. The word “best” depends on context. It might mean highest, lowest, fastest, cheapest, safest, strongest, or most efficient.

In business, a profit function may be quadratic because raising price increases revenue per item but decreases the number of buyers. The vertex gives the price that maximizes profit under the assumptions of the model. In engineering, material strength or cost may depend on dimensions in a curved way. The vertex can help identify a minimum-cost design or maximum-performance setting. In sports, height, distance, and speed models often involve quadratics. The vertex locates the highest point of a jump or throw. In computer graphics, parabolic paths are used for animation, simulations, and game physics.

The technique also represents fairness in comparison. Suppose two students are comparing two expressions and one is written in standard form while the other is written in vertex form. Without converting, they may not see that the expressions define the same function or have the same maximum. Completing the square gives a common language for comparing quadratics.

It also helps students see why some problems have no real solution. If a quadratic equation becomes \((x - 2)^2 + 5 = 0\), then it has no real solutions because a square plus 5 cannot be zero. This connects completing the square to the complex number system, which Math II develops more fully. The same form that reveals maxima and minima also reveals whether a quadratic can cross the x-axis.

Where this fits in the big map of mathematics

In the full map of mathematics, completing the square is a central intersection. It connects algebraic rewriting to geometric structure. It connects equations to functions. It connects graph features to symbolic form. It prepares students for conic sections, where completing the square reveals centers and radii. It prepares students for optimization, where identifying maximum and minimum values becomes a major theme. It even prepares students for calculus, where finding extrema becomes formalized through derivatives.

Students should see three major quadratic forms as three different windows. Standard form shows coefficients and is often easiest for arithmetic. Factored form shows zeros and x-intercepts. Vertex form shows the turning point, symmetry, and extreme value. A strong student does not ask which form is “the right form” in general. A strong student asks which form answers the current question.

This objective is therefore not just about one technique. It is about representational intelligence. Students learn to choose and produce equivalent forms in order to reveal meaning. That is one of the most important habits in all advanced mathematics.

Common student traps and how to avoid them

One trap is forgetting to factor out the leading coefficient before completing the square. In \(3x^2 + 12x + 5\), students sometimes add 36 because half of 12 is 6. But the correct move is to factor first: \(3(x^2 + 4x) + 5\), then add and subtract 4 inside the parentheses. The leading coefficient changes the bookkeeping.

A second trap is adding a number without subtracting it back. Rewriting must preserve equivalence. If students add the missing square, they must compensate so the expression’s value does not change.

A third trap is mixing up the sign of the vertex. In \((x - h)^2 + k\), the vertex is \((h, k)\), not \((-h, k)\). The expression \((x - 3)^2 + 7\) has vertex \((3, 7)\) because the square is zero when \(x = 3\).

A fourth trap is treating the vertex as always a maximum. Upward-opening parabolas have minimums. Downward-opening parabolas have maximums. The sign of the leading coefficient decides.

Problem Library

Problems in the App From This Objective

168 problems across 12 archetypes in the app.

create a perfect-square trinomial.
15 problems Warmup Practice Mixed Review Assessment
Problem 1

Convert monic quadratic x^2+8x+3 to vertex form by completing the square.

Problem 2

Convert monic quadratic x^2-10x+6 to vertex form by completing the square.

Problem 3

Convert monic quadratic x^2+3x-2 to vertex form by completing the square.

Open in simulator
Problem 4

Convert monic quadratic x^2-7x+1 to vertex form by completing the square.

Problem 5

Convert monic quadratic x^2+2x+5 to vertex form by completing the square.

Problem 6

Convert monic quadratic x^2-4x-1 to vertex form by completing the square.

Problem 7

Convert monic quadratic x^2+6x to vertex form by completing the square.

Problem 8

Convert monic quadratic x^2-x+1 to vertex form by completing the square.

Problem 9

Convert monic quadratic x^2+5x-3 to vertex form by completing the square.

Problem 10

Convert monic quadratic x^2-9x+2 to vertex form by completing the square.

Problem 11

Convert monic quadratic x^2+12x+30 to vertex form by completing the square.

Problem 12

Convert monic quadratic x^2-14x+50 to vertex form by completing the square.

Problem 13

Convert monic quadratic x^2+x to vertex form by completing the square.

Problem 14

Convert monic quadratic x^2-2x to vertex form by completing the square.

Problem 15

Convert monic quadratic x^2+10x-1 to vertex form by completing the square.

factor leading coefficient before completing square.
15 problems Warmup Practice Mixed Review Assessment
Problem 16

Convert non-monic quadratic 2x^2+12x+5 to vertex form.

Problem 17

Convert non-monic quadratic 3x^2-18x+4 to vertex form.

Problem 18

Convert non-monic quadratic -x^2+6x+1 to vertex form.

Problem 19

Convert non-monic quadratic -2x^2-8x+7 to vertex form.

Problem 20

Convert non-monic quadratic 4x^2+16x+3 to vertex form.

Problem 21

Convert non-monic quadratic 5x^2-20x+1 to vertex form.

Problem 22

Convert non-monic quadratic -3x^2+12x-2 to vertex form.

Problem 23

Convert non-monic quadratic -4x^2-24x+9 to vertex form.

Problem 24

Convert non-monic quadratic 2x^2+8x-1 to vertex form.

Problem 25

Convert non-monic quadratic 3x^2-6x+10 to vertex form.

Problem 26

Convert non-monic quadratic -2x^2+16x-3 to vertex form.

Problem 27

Convert non-monic quadratic 5x^2+10x+2 to vertex form.

Open in simulator
Problem 28

Convert non-monic quadratic -3x^2-6x+5 to vertex form.

Problem 29

Convert non-monic quadratic 4x^2-8x+1 to vertex form.

Problem 30

Convert non-monic quadratic -5x^2+30x-1 to vertex form.

read `(h,k)` from `a(x-h)^2+k`.
15 problems Warmup Practice Mixed Review Assessment
Problem 31

Identify the vertex of quadratic f(x)=3(x-2)^2+5 from completed-square form.

Problem 32

Identify the vertex of quadratic g(x)=-(x+4)^2-1 from completed-square form.

Problem 33

Identify the vertex of quadratic h(x)=0.5(x-7)^2+9 from completed-square form.

Problem 34

Identify the vertex of quadratic p(x)=-2(x+3/2)^2+4 from completed-square form.

Problem 35

Identify the vertex of quadratic f(x)=(x-1)^2+2 from completed-square form.

Problem 36

Identify the vertex of quadratic f(x)=(x+5)^2-3 from completed-square form.

Problem 37

Identify the vertex of quadratic f(x)=-2(x-6)^2-7 from completed-square form.

Problem 38

Identify the vertex of quadratic f(x)=1/2(x+1/3)^2+8 from completed-square form.

Problem 39

Identify the vertex of quadratic f(x)=-3/4(x-1/5)^2-9 from completed-square form.

Problem 40

Identify the vertex of quadratic f(x)=1.5(x-2.1)^2+3.2 from completed-square form.

Problem 41

Identify the vertex of quadratic f(x)=-0.5(x+1.1)^2-2.2 from completed-square form.

Problem 42

Identify the vertex of quadratic f(x)=4x^2+10 from completed-square form.

Problem 43

Identify the vertex of quadratic f(x)=5(x+6)^2 from completed-square form.

Problem 44

Identify the vertex of quadratic f(x)=x^2 from completed-square form.

Problem 45

Identify the vertex of quadratic f(x)=-3(x-1/4)^2+1/2 from completed-square form.

Open in simulator
use sign of leading coefficient.
15 problems Warmup Practice Mixed Review Assessment
Problem 46

Determine whether f(x)=2(x-1)^2-3 has a maximum or minimum and explain why.

Open in simulator
Problem 47

Determine whether g(x)=-(x+5)^2+8 has a maximum or minimum and explain why.

Problem 48

Determine whether h(x)=0.25x^2+6x+1 has a maximum or minimum and explain why.

Problem 49

Determine whether p(x)=-4x^2+12x-7 has a maximum or minimum and explain why.

Problem 50

Determine whether y = 3(x+2)^2 + 1 has a maximum or minimum and explain why.

Problem 51

Determine whether y = -0.5(x-4)^2 - 2 has a maximum or minimum and explain why.

Problem 52

Determine whether f(x) = x^2 - 5x + 6 has a maximum or minimum and explain why.

Problem 53

Determine whether f(x) = -2x^2 + 3x - 1 has a maximum or minimum and explain why.

Problem 54

Determine whether y = (x-1)(x-3) has a maximum or minimum and explain why.

Problem 55

Determine whether y = -(x+2)(x-4) has a maximum or minimum and explain why.

Problem 56

Determine whether f(x) = (1/3)x^2 - 2x + 5 has a maximum or minimum and explain why.

Problem 57

Determine whether g(x) = -1.5x^2 + 7x - 10 has a maximum or minimum and explain why.

Problem 58

Determine whether h(x) = (1/4)(x-3)^2 + 2 has a maximum or minimum and explain why.

Problem 59

Determine whether p(x) = -0.75(x+1)^2 - 6 has a maximum or minimum and explain why.

Problem 60

Determine whether y = 5x^2 - 1 has a maximum or minimum and explain why.

identify k-value and interpret.
12 problems Warmup Practice Mixed Review Assessment
Problem 61

Find the maximum or minimum value of f(x)=(x-6)^2+2 from vertex form.

Problem 62

Find the maximum or minimum value of g(x)=-3(x+1)^2+12 from vertex form.

Problem 63

Find the maximum or minimum value of h(x)=0.5(x-4)^2-8 from vertex form.

Open in simulator
Problem 64

Find the maximum or minimum value of p(x)=-(x-9)^2 from vertex form.

Problem 65

Find the maximum or minimum value of y=2(x-3)^2+5 from vertex form.

Problem 66

Find the maximum or minimum value of y=-0.5(x+2)^2+7 from vertex form.

Problem 67

Find the maximum or minimum value of y=(x+5)^2-1 from vertex form.

Problem 68

Find the maximum or minimum value of y=-4(x-1)^2-10 from vertex form.

Problem 69

Find the maximum or minimum value of y=3x^2+6 from vertex form.

Problem 70

Find the maximum or minimum value of y=-x^2-3 from vertex form.

Problem 71

Find the maximum or minimum value of y=0.25(x-7)^2 from vertex form.

Problem 72

Find the maximum or minimum value of y=-2(x+8)^2 from vertex form.

model and interpret vertex as optimum.
15 problems Warmup Practice Mixed Review Assessment
Problem 73

Use completing the square to solve the optimization context: Revenue is R(p)=-p^2+18p. Find the price that maximizes revenue and the maximum revenue.

Problem 74

Use completing the square to solve the optimization context: Area is A(x)=-2x^2+20x. Find the maximum area.

Problem 75

Use completing the square to solve the optimization context: Height is h(t)=-16t^2+64t+5. Find maximum height.

Problem 76

Use completing the square to solve the optimization context: Cost is C(x)=x^2-12x+45. Find minimum cost.

Problem 77

Use completing the square to solve the optimization context: Profit is P(x)=-3x^2+30x-50. Find the maximum profit.

Problem 78

Use completing the square to solve the optimization context: A ball is thrown with height H(t)=-5t^2+40t+10. Find the maximum height.

Problem 79

Use completing the square to solve the optimization context: A farmer wants to fence a rectangular area next to a river. He has 100 meters of fence. Let x be the side perpendicular to the river. The area is A(x)=-2x^2+100x. Find the maximum area.

Problem 80

Use completing the square to solve the optimization context: The cost to produce x units is C(x)=2x^2-24x+80. Find the minimum cost.

Problem 81

Use completing the square to solve the optimization context: Revenue is R(p)=-0.5p^2+10p. Find the price that maximizes revenue and the maximum revenue.

Problem 82

Use completing the square to solve the optimization context: The sum of two numbers is 20. Let one number be x. Their product is P(x)=-x^2+20x. Find the maximum product.

Problem 83

Use completing the square to solve the optimization context: Find the minimum value of the function f(x)=3x^2-18x+27.

Problem 84

Use completing the square to solve the optimization context: Find the maximum value of the function g(x)=-4x^2-16x-10.

Problem 85

Use completing the square to solve the optimization context: A company's daily production cost is modeled by C(q)=0.1q^2-10q+500, where q is the number of units produced. Find the minimum daily cost.

Problem 86

Use completing the square to solve the optimization context: The path of a rocket is given by y=-0.01x^2+x+10, where y is the height and x is the horizontal distance. Find the maximum height the rocket reaches.

Problem 87

Use completing the square to solve the optimization context: The squared distance from a point (x, y) on the line y=x+1 to the point (2,0) is given by D(x)=2x^2-2x+5. Find the minimum squared distance.

Open in simulator
explain which features each form reveals.
15 problems Warmup Practice Mixed Review Assessment
Problem 88

Compare what standard form x^2-6x+5 and vertex form (x-3)^2-4 reveal.

Problem 89

Compare what standard form -2x^2+8x+1 and vertex form -2(x-2)^2+9 reveal.

Problem 90

Compare what standard form x^2+4x+7 and vertex form (x+2)^2+3 reveal.

Problem 91

Compare what standard form 3x^2-12x+6 and vertex form 3(x-2)^2-6 reveal.

Problem 92

Compare what standard form x^2-4x and vertex form (x-2)^2-4 reveal.

Problem 93

Compare what standard form -x^2+6x and vertex form -(x-3)^2+9 reveal.

Problem 94

Compare what standard form 2x^2+5 and vertex form 2x^2+5 reveal.

Problem 95

Compare what standard form 0.5x^2+2x+1 and vertex form 0.5(x+2)^2-1 reveal.

Problem 96

Compare what standard form -3x^2-6x-2 and vertex form -3(x+1)^2+1 reveal.

Problem 97

Compare what standard form x^2-8x+16 and vertex form (x-4)^2 reveal.

Problem 98

Compare what standard form -x^2+3 and vertex form -x^2+3 reveal.

Open in simulator
Problem 99

Compare what standard form 2x^2+4x+3 and vertex form 2(x+1)^2+1 reveal.

Problem 100

Compare what standard form x^2+2x-3 and vertex form (x+1)^2-4 reveal.

Problem 101

Compare what standard form 5x^2-20x+18 and vertex form 5(x-2)^2-2 reveal.

Problem 102

Compare what standard form -0.5x^2-3x-4.5 and vertex form -0.5(x+3)^2 reveal.

identify `x=h`.
15 problems Warmup Practice Mixed Review Assessment
Problem 103

Find the axis of symmetry of f(x)=(x-4)^2+1.

Problem 104

Find the axis of symmetry of g(x)=-2(x+3)^2+8.

Open in simulator
Problem 105

Find the axis of symmetry of h(x)=x^2-10x+7.

Problem 106

Find the axis of symmetry of p(x)=2x^2+6x-1.

Problem 107

Find the axis of symmetry of y=(x+1)^2-5.

Problem 108

Find the axis of symmetry of y=-(x-7)^2+2.

Problem 109

Find the axis of symmetry of y=3x^2+4.

Problem 110

Find the axis of symmetry of y=-0.5(x+2.5)^2-10.

Problem 111

Find the axis of symmetry of y=(x-1/2)^2+3.

Problem 112

Find the axis of symmetry of y=x^2+4x+3.

Problem 113

Find the axis of symmetry of y=-x^2+8x-5.

Problem 114

Find the axis of symmetry of y=3x^2-12x+1.

Problem 115

Find the axis of symmetry of y=-2x^2-10x+7.

Problem 116

Find the axis of symmetry of y=0.5x^2+3x-2.

Problem 117

Find the axis of symmetry of y=4x^2+5.

plot vertex, axis, and symmetric points.
12 problems Warmup Practice Mixed Review Assessment
Problem 118

Use completed-square form f(x)=(x-2)^2-3 to list key features needed for a sketch.

Problem 119

Use completed-square form g(x)=-(x+1)^2+4 to list key features needed for a sketch.

Problem 120

Use completed-square form h(x)=2(x-3)^2+1 to list key features needed for a sketch.

Problem 121

Use completed-square form p(x)=-0.5(x-4)^2+2 to list key features needed for a sketch.

Problem 122

Use completed-square form y=(x+3)^2-1 to list key features needed for a sketch.

Problem 123

Use completed-square form y=-(x-1)^2+5 to list key features needed for a sketch.

Problem 124

Use completed-square form y=3(x+2)^2-4 to list key features needed for a sketch.

Problem 125

Use completed-square form y=-2(x-5)^2+3 to list key features needed for a sketch.

Problem 126

Use completed-square form y=0.5x^2 to list key features needed for a sketch.

Problem 127

Use completed-square form y=-0.5(x+4)^2-2 to list key features needed for a sketch.

Problem 128

Use completed-square form y=(x-10)^2+7 to list key features needed for a sketch.

Open in simulator
Problem 129

Use completed-square form y=-(x+6)^2-5 to list key features needed for a sketch.

use vertex and opening direction.
15 problems Warmup Practice Mixed Review Assessment
Problem 130

Determine the range of quadratic f(x)=(x-3)^2+2.

Open in simulator
Problem 131

Determine the range of quadratic g(x)=-4(x+1)^2+9.

Problem 132

Determine the range of quadratic h(x)=0.5(x-6)^2-7.

Problem 133

Determine the range of quadratic p(x)=-(x-2)^2.

Problem 134

Determine the range of quadratic y = (x+5)^2 + 1.

Problem 135

Determine the range of quadratic y = -2(x-1)^2 - 3.

Problem 136

Determine the range of quadratic y = 3(x-7)^2.

Problem 137

Determine the range of quadratic y = -0.5(x+4)^2 + 6.

Problem 138

Determine the range of quadratic y = 1/3(x)^2 - 5.

Problem 139

Determine the range of quadratic y = -1/4(x+2)^2.

Problem 140

Determine the range of quadratic y = 5(x-10)^2 + 15.

Problem 141

Determine the range of quadratic y = -10(x+8)^2 - 12.

Problem 142

Determine the range of quadratic y = 1.5(x-0.5)^2 + 2.5.

Problem 143

Determine the range of quadratic y = -2.5(x+1.5)^2 - 0.5.

Problem 144

Determine the range of quadratic y = 2x^2.

reason that squared term is nonnegative.
12 problems Warmup Practice Mixed Review Assessment
Problem 145

Explain why completed-square form f(x)=(x-5)^2+3 reveals an extreme value.

Problem 146

Explain why completed-square form g(x)=-(x+2)^2+10 reveals an extreme value.

Problem 147

Explain why completed-square form h(x)=4(x-1)^2-8 reveals an extreme value.

Problem 148

Explain why completed-square form p(x)=-2(x-7)^2+6 reveals an extreme value.

Problem 149

Explain why completed-square form f(x)=(x+1)^2-4 reveals an extreme value.

Problem 150

Explain why completed-square form g(x)=-(x-3)^2+7 reveals an extreme value.

Problem 151

Explain why completed-square form h(x)=3(x+6)^2+1 reveals an extreme value.

Problem 152

Explain why completed-square form p(x)=-0.5x^2-2 reveals an extreme value.

Problem 153

Explain why completed-square form q(x)=(1/2)(x-4)^2 reveals an extreme value.

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Problem 154

Explain why completed-square form r(x)=-(3/4)(x+9)^2+5 reveals an extreme value.

Problem 155

Explain why completed-square form s(x)=5(x+10)^2-1 reveals an extreme value.

Problem 156

Explain why completed-square form t(x)=-10(x-1)^2+100 reveals an extreme value.

catch half-b, factor-a, sign, and vertex-reading mistakes.
12 problems Warmup Practice Mixed Review Assessment
Problem 157

Correct the completed-square extrema error in x^2+8x+1=(x+8)^2+1, so the minimum is 1.

Problem 158

Correct the completed-square extrema error in -x^2+6x+2=-(x+3)^2+2, so the maximum is 2.

Problem 159

Correct the completed-square extrema error in 2x^2-8x+5=2(x-4)^2+5, so vertex is (4,5).

Problem 160

Correct the completed-square extrema error in (x-1)^2+7 has maximum 7.

Problem 161

Correct the completed-square extrema error in x^2 - 6x + 2 = (x-3)^2 + 2, so the minimum is 2.

Problem 162

Correct the completed-square extrema error in x^2 + 10x + 1 = (x-5)^2 + 1, so the minimum is 1.

Problem 163

Correct the completed-square extrema error in -x^2 - 4x + 3 = -(x-2)^2 + 3, so the minimum is 3.

Problem 164

Correct the completed-square extrema error in 3x^2 + 6x + 4 = 3(x+1)^2 + 4, so the minimum is 4.

Problem 165

Correct the completed-square extrema error in (x+2)^2 - 5, so the vertex is (2, -5).

Problem 166

Correct the completed-square extrema error in x^2 - 8x + 10 = (x-4)^2 + 10, so the minimum is 10.

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Problem 167

Correct the completed-square extrema error in x^2 + 3x + 1 = (x+3/2)^2 + 1, so the minimum is 1.

Problem 168

Correct the completed-square extrema error in -2x^2 + 12x - 7 = -2(x-3)^2 - 7, so the maximum is -7.