College Algebra Quiz: Vertex Form And Completing The Square

10 questions · exam conditions

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Vertex Form And Completing The SquareQuestion 1 of 10

A quadratic function $f(x) = x^2 + bx + c$ has its vertex at $(4, -3)$ . What is the value of $b$ ?

b = 4

b = 8

b = -4

b = -8

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College Algebra Quiz

College Algebra Quiz: Vertex Form And Completing The Square

Practice Vertex Form And Completing The Square in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Vertex Form And Completing The Square, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A quadratic function $f(x) = x^2 + bx + c$ has its vertex at $(4, -3)$ . What is the value of $b$ ?

$b = 4$
$b = 8$
$b = -4$
$b = -8$ (correct answer)

Explanation: When you encounter a quadratic function with a given vertex, you're working with the relationship between the standard form

f(x) = x^2 + bx + c

and the vertex form. The key insight is that for any quadratic

f(x) = ax^2 + bx + c

, the x-coordinate of the vertex is always

x = -\frac{b}{2a}

. Since our function is

f(x) = x^2 + bx + c

, we have

a = 1

. Given that the vertex is at

(4, -3)

, the x-coordinate of the vertex is 4. Using the vertex formula:

4 = -\frac{b}{2(1)} = -\frac{b}{2}

Solving for

b

8 = -b

, so

b = -8

. Let's examine why the other answers are incorrect. Answer choice A (

b = 4

) occurs if you mistakenly think the x-coordinate of the vertex equals

b

directly. Answer choice B (

b = 8

) happens if you forget the negative sign in the vertex formula, setting

4 = \frac{b}{2}

. Answer choice C (

b = -4

) results from incorrectly using

4 = -\frac{b}{2}

but solving as

4 \times 2 = -b

, giving

8 = -b

and then forgetting to apply the negative properly. The correct answer is D:

b = -8

. Study tip: Memorize that for

f(x) = ax^2 + bx + c

, the vertex occurs at

x = -\frac{b}{2a}

. Always be careful with the negative sign—it's the most common source of error in vertex problems.

Question 2

A quadratic function has vertex at $(-3, 5)$ and passes through the point $(1, -11)$ . What is the value of $a$ in the vertex form $f(x) = a(x - h)^2 + k$ ?

$a = -1$ (correct answer)
$a = -\frac{1}{4}$
$a = \frac{1}{4}$
$a = 1$

Explanation: With vertex at

(-3, 5)

, the vertex form is

f(x) = a(x - (-3))^2 + 5 = a(x + 3)^2 + 5

. Using the point

(1, -11)

-11 = a(1 + 3)^2 + 5 = a(16) + 5 = 16a + 5

. Solving:

-16 = 16a

, so

a = -1

. Choice B incorrectly uses

(h, k) = (3, 5)

. Choice C gets the sign wrong. Choice D ignores the negative coefficient needed for the parabola to open downward.

Question 3

The vertex form of a parabola is $g(x) = \frac{1}{2}(x - 5)^2 + 7$ . If this function is written in standard form $g(x) = ax^2 + bx + c$ , what is the value of $c$ ?

$c = 32$
$c = 7$
$c = 12$
$c = 19.5$ (correct answer)

Explanation: When you encounter a question asking you to convert from vertex form to standard form, you're working with two different ways to express the same quadratic function. The vertex form

g(x) = a(x - h)^2 + k

highlights the vertex at

(h, k)

, while standard form

g(x) = ax^2 + bx + c

shows the y-intercept as

c

. To find

c

, you need to expand the vertex form. Starting with

g(x) = \frac{1}{2}(x - 5)^2 + 7

, first expand

(x - 5)^2 = x^2 - 10x + 25

. Then multiply by

\frac{1}{2}

\frac{1}{2}(x^2 - 10x + 25) = \frac{1}{2}x^2 - 5x + \frac{25}{2}

. Finally, add 7:

g(x) = \frac{1}{2}x^2 - 5x + \frac{25}{2} + 7 = \frac{1}{2}x^2 - 5x + \frac{25}{2} + \frac{14}{2} = \frac{1}{2}x^2 - 5x + \frac{39}{2}

. Therefore,

c = \frac{39}{2} = 19.5

, making D correct. Choice A (

c = 32

) likely comes from incorrectly computing

25 + 7 = 32

without considering the

\frac{1}{2}

coefficient. Choice B (

c = 7

) represents the

k

-value from vertex form, confusing the y-coordinate of the vertex with the y-intercept. Choice C (

c = 12

) might result from calculation errors in the expansion process. Remember: the constant term

c

in standard form is the y-intercept, which you can also find by evaluating

g(0)

. This provides a useful check for your algebra.

Question 4

When the expression $-2x^2 + 12x - 7$ is written in vertex form $a(x - h)^2 + k$ , what is the value of $k$ ?

$k = -7$
$k = 11$ (correct answer)
$k = 25$
$k = -25$

Explanation: When you encounter a quadratic expression that needs to be converted to vertex form, you're working with completing the square. Vertex form

a(x - h)^2 + k

reveals the vertex of the parabola at point

(h, k)

, where

k

represents the minimum or maximum value of the function. To convert

-2x^2 + 12x - 7

to vertex form, start by factoring out the coefficient of

x^2

from the first two terms:

-2(x^2 - 6x) - 7

. Now complete the square inside the parentheses. Take half of the coefficient of

x

(which is

-6

), square it:

(-3)^2 = 9

. Add and subtract this inside the parentheses:

-2(x^2 - 6x + 9 - 9) - 7

. This gives us

-2((x - 3)^2 - 9) - 7

. Distribute the

-2

-2(x - 3)^2 + 18 - 7 = -2(x - 3)^2 + 11

. Therefore,

k = 11

. Answer choice A)

k = -7

is the original constant term, but this doesn't account for the adjustments made during completing the square. Answer choice C)

k = 25

likely comes from incorrectly adding

18

instead of subtracting when distributing. Answer choice D)

k = -25

represents a sign error combined with the mistake in choice C. Remember: when completing the square, carefully track your arithmetic, especially when distributing negative coefficients. The

k

value in vertex form is never simply the original constant term—it's the result of the entire transformation process.

Question 5

Which expression represents the result of completing the square for $2x^2 - 12x + 7$ ?

$2(x - 3)^2 - 11$ (correct answer)
$2(x - 6)^2 - 65$
$(2x - 6)^2 - 29$
$2(x - 3)^2 + 25$

Explanation: First factor out the coefficient of

x^2

2x^2 - 12x + 7 = 2(x^2 - 6x) + 7

. Complete the square inside:

x^2 - 6x = (x - 3)^2 - 9

. So

2(x^2 - 6x) + 7 = 2((x - 3)^2 - 9) + 7 = 2(x - 3)^2 - 18 + 7 = 2(x - 3)^2 - 11

. Choice B uses wrong value for

h

. Choice C doesn't factor out the 2 correctly. Choice D has wrong sign for the constant term.

Question 6

A parabola in vertex form $f(x) = -2(x + 1)^2 + 8$ is transformed by shifting it 3 units right and 2 units down. What is the vertex of the transformed function?

$(2, 6)$ (correct answer)
$(-4, 10)$
$(4, -2)$
$(1, -2)$

Explanation: The original vertex is at

(-1, 8)

from the form

f(x) = -2(x - (-1))^2 + 8

. Shifting 3 units right means adding 3 to the x-coordinate:

-1 + 3 = 2

. Shifting 2 units down means subtracting 2 from the y-coordinate:

8 - 2 = 6

. The new vertex is

(2, 6)

. Choice B shifts left and up instead. Choice C uses wrong original vertex. Choice D only applies one transformation correctly.

Question 7

To complete the square for $3x^2 + 18x - 5$ , what value must be added and subtracted inside the parentheses after factoring out the leading coefficient?

$9$ (correct answer)
$3$
$27$
$6$

Explanation: After factoring out 3:

3x^2 + 18x - 5 = 3(x^2 + 6x) - 5

. To complete the square for

x^2 + 6x

, take half of the coefficient of

x

and square it:

\left(\frac{6}{2}\right)^2 = 3^2 = 9

. So we add and subtract 9:

x^2 + 6x = (x + 3)^2 - 9

. Choice B uses half the coefficient without squaring. Choice C multiplies by the leading coefficient incorrectly. Choice D uses the coefficient of

x

directly.

Question 8

A parabola has equation $y = -x^2 + 8x + 5$ . What are the coordinates of its vertex?

$(-4, -43)$
$(8, 5)$
$(4, 21)$ (correct answer)
$(4, 5)$

Explanation: When you encounter a parabola in standard form

y = ax^2 + bx + c

, finding the vertex requires identifying the point where the parabola reaches its maximum or minimum value. For any parabola, the vertex occurs at

x = -\frac{b}{2a}

. In the equation

y = -x^2 + 8x + 5

, we have

a = -1

b = 8

, and

c = 5

. The x-coordinate of the vertex is:

x = -\frac{8}{2(-1)} = -\frac{8}{-2} = 4

To find the y-coordinate, substitute

x = 4

back into the original equation:

y = -(4)^2 + 8(4) + 5 = -16 + 32 + 5 = 21

Therefore, the vertex is at

(4, 21)

, which is answer choice C. Looking at the wrong answers: Choice A

(-4, -43)

likely results from sign errors in the vertex formula or substitution. Choice B

(8, 5)

incorrectly uses the coefficient

b = 8

as the x-coordinate and the constant term

c = 5

as the y-coordinate. Choice D

(4, 5)

gets the correct x-coordinate but mistakenly uses the constant term

c = 5

instead of properly substituting to find the y-coordinate. Remember: For vertex problems, always use the formula

x = -\frac{b}{2a}

to find the x-coordinate, then substitute back into the original equation to find y. Don't assume the coefficients in the equation directly give you coordinate values.

Question 9

Which of the following represents $4x^2 - 24x + 11$ after completing the square?

$(4x - 12)^2 - 133$
$4(x - 6)^2 - 133$
$4(x - 3)^2 - 25$ (correct answer)
$4(x - 3)^2 + 47$

Explanation: Completing the square is a technique for rewriting quadratic expressions in vertex form, which reveals important properties like the vertex of a parabola. When you see a quadratic like

4x^2 - 24x + 11

, you're transforming it into the form

a(x - h)^2 + k

. Start by factoring out the coefficient of

x^2

from the first two terms:

4x^2 - 24x + 11 = 4(x^2 - 6x) + 11

. Now focus on completing the square inside the parentheses. Take half of the coefficient of

x

(which is

-6

), square it:

(-6/2)^2 = (-3)^2 = 9

. Add and subtract this value:

4(x^2 - 6x + 9 - 9) + 11 = 4((x^2 - 6x + 9) - 9) + 11

. The expression

x^2 - 6x + 9

is a perfect square:

(x - 3)^2

. So you have:

4((x - 3)^2 - 9) + 11 = 4(x - 3)^2 - 36 + 11 = 4(x - 3)^2 - 25

. This matches choice C. Choice A incorrectly keeps the 4 inside the squared term and miscalculates the constant. Choice B has the wrong value inside the parentheses (6 instead of 3) and wrong constant term. Choice D has the correct squared term but adds 47 instead of subtracting 25, likely from sign errors in the arithmetic. Remember: when completing the square with a leading coefficient other than 1, always factor it out first, complete the square on the simpler expression, then distribute carefully to avoid sign and arithmetic errors.

Question 10

A quadratic function in vertex form is $f(x) = 3(x + 2)^2 - 12$ . If the graph is reflected across the x-axis and then shifted up 5 units, what is the vertex of the resulting function?

$(-2, -7)$
$(-2, 17)$ (correct answer)
$(2, -7)$
$(-2, 7)$

Explanation: When working with transformations of quadratic functions, you need to track how each transformation affects both the shape and position of the parabola. The original function

f(x) = 3(x + 2)^2 - 12

is in vertex form

a(x - h)^2 + k

, so its vertex is at

(-2, -12)

. Let's apply the transformations step by step. First, reflecting across the x-axis changes the sign of the entire function, giving us

g(x) = -3(x + 2)^2 + 12

. This reflection flips the parabola upside down and moves the vertex from

(-2, -12)

(-2, 12)

because the y-coordinate becomes its opposite. Next, shifting up 5 units means adding 5 to the entire function:

h(x) = -3(x + 2)^2 + 12 + 5 = -3(x + 2)^2 + 17

. This moves the vertex from

(-2, 12)

(-2, 17)

, which is choice B. Choice A

(-2, -7)

represents a common error where you might have applied the transformations incorrectly or forgotten that reflecting across the x-axis changes the sign of the y-coordinate. Choice C

(2, -7)

incorrectly changes the x-coordinate of the vertex, but horizontal shifts don't affect the vertex's x-position when there's no horizontal transformation applied. Choice D

(-2, 7)

likely comes from miscalculating the final vertical position after the transformations. Remember: reflections across the x-axis negate all y-values, while vertical shifts add directly to the y-coordinate. Always apply transformations in order and track how each affects the vertex coordinates separately.