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Precalculus : Express a Quadratic Function in Vertex Form

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Example Question #1 : Express A Quadratic Function In Vertex Form

Which of the following is the given quadratic equation in vertex form?

$y=5x^{2}+20x+27$

Possible Answers:

$y=5(x+2)^{2}+23$

$y=5(x-4)^{2}+3$

$y=5(x+2)^{2}+27$

$y=5(x+4)^{2}+7$

$y=5(x+2)^{2}+7$

Correct answer:

$y=5(x+2)^{2}+7$

Explanation:

To solve for the vertex form, we must start by completing the square:

y=5x^2+20x+27

$y+\; \; \; \; =5(x^2+4x+\; \; \; \; )+27$

y+5(4) =5(x^2+4x+4)+27

$y+20=5(x+2)^{2}+27$

$y=5(x+2)^{2}+7$

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Example Question #2 : Express A Quadratic Function In Vertex Form

Which of the following is the appropriate vertex form of the following quadratic equation?

$y=3x^{2}+15x+32$

Possible Answers:

$y=3(x+5)^{2}+17$

$y=3(x+\frac{5}{2})^{2}+\frac{53}{4}$

$y=3(x+\frac{2}{5})^{2}+20$

$y=3(x+\frac{5}{2})^{2}+\frac{103}{4}$

$y=3(x+5)^{2}+29$

Correct answer:

$y=3(x+\frac{5}{2})^{2}+\frac{53}{4}$

Explanation:

This process outlines how to convert a quadratic function to vertex form:

$y=3x^{2}+15x+32$

$y+\; \; \; \; =3(x^{2}+5x+\; \; \; \; )+32$

$y+3(\frac{5}{2})^{2} =3(x^{2}+5x+(\frac{5}{2})^{2} )+32$

$y+\frac{75}{4} =3(x^{2}+5x+\frac{25}{4})+32$

$y=3(x^{2}+5x+\frac{25}{4})+32-\frac{75}{4}$

$y=3(x+\frac{5}{2})^{2}+\frac{128}{4}-\frac{75}{4}$

$y=3(x+\frac{5}{2})^{2}+\frac{53}{4}$

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Example Question #3 : Express A Quadratic Function In Vertex Form

Give the coordinate pair of the vertex of this quadratic function x^2+2x-6 .

Possible Answers:

(h,k)=(-1,7)

(h,k)=(-1,-7)

None of the other answers.

(h,k)=(1,7)

(h,k)=(1,-7)

Correct answer:

(h,k)=(-1,-7)

Explanation:

Expressing quadratic functions in the vertex form is basically just changing the format of the equation to give us different information, namely the vertex. In order for us to change the function into this format we must have it in standard form ax^2+bx+c . After that, our goal is to change the function into the form y=a(x-h)^2+k . We do so as follows:

ax^2+bx+c=0

subtract the constant over to the other side ax^2+bx=-c

halve the b term, square it, and add to both sides. $ax^2+bx+\left(\frac{b}{2}\right)^2=-c+\left(\frac{b}{2}\right)^2$

Now factor the left side. $\left(x+\frac{b}{2}\right)^2=-c+\left(\frac{b}{2}\right)^2$

now simplify the right side and move that number back over to the left side and you will be left with a(x-h)^2+k . I recommend looking up an example with numbers before you begin or at least recognizing that the fractions will end up being whole numbers in most problems. Below is specific explanation of the problem at hand. Try to use the generic equation to find the answer before following the step by step approach below.

x^2+2x-6=0

move the constant over

x^2+2x=6

halve the b term and add to both sides

x^2+2x+1=6+1

factor the left side and simplify the right

(x+1)^2=7

move the constant over to achieve vertex form

(x+1)^2-7 is the final answer with vertex at (-1,-7). Note that the formula is y=a(x-h)^2+k .

try this shortcut after you have mastered the steps: (h,k)= (-b/2a, 4ac-b^2/4a) . Make sure you recognize that this formula gives you an x and y coordinate for the vertex and that each coordinate of the pair is fraction in the formula. This will give you the vertex of the equation if it is in standard form. However, don't rely on this as completing the square is also a method for finding the roots. So you need to know both methods before you cut the corner.

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Example Question #4 : Express A Quadratic Function In Vertex Form

Express the following quadratic equation in vertex form.

y=2x^2-28x+108

Possible Answers:

y=2(x-10)^2+7

y=(x-14)^2+108

y=2(x-7)^2+10

y=(x-5)^2+7

Correct answer:

y=2(x-7)^2+10

Explanation:

To get the equation into vertex form, we factor the largest constant from the terms with a degree of greater than or equal to 1.

y=2x^2-28x+108

y=2(x^2-14x)+108

We then complete the square by following these steps

finding half of the coefficients of the term
squaring that result
and then adding that square to the expression for .

Keep in mind that what is done on one side of the equation must be done on the other.

y+2*7^2=2(x^2-14x+7^2)+108

y+98=2(x^2-14x+49)+108

And factoring the quadratic polynomial of x

y+98=2(x-7)^2+108

we get

y=2(x-7)^2+10

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Precalculus : Express a Quadratic Function in Vertex Form

Study concepts, example questions & explanations for Precalculus

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Example Question #1 : Express A Quadratic Function In Vertex Form

Example Question #2 : Express A Quadratic Function In Vertex Form

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