How do you write $y={x}^{2}+16x+14$ in vertex form?

compagnia04
2021-12-16
Answered

How do you write $y={x}^{2}+16x+14$ in vertex form?

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Jordan Mitchell

Answered 2021-12-17
Author has **31** answers

By completing the square:

$y={x}^{2}+16x+{\left(\frac{16}{2}\right)}^{2}-{\left(\frac{16}{2}\right)}^{2}+14$

$y={(x+8)}^{2}-64+14$

$y={(x+8)}^{2}-50$

vicki331g8

Answered 2021-12-18
Author has **37** answers

Complete the square for ${x}^{2}+16x+14$

Use the form$a{x}^{2}+bx+c$ , to find the values of $a,b,$ and c.

Consider the vertex form of a parabola.

$a{(x+d)}^{2}+e$

Substitute the values of a and b into the formula$d=\frac{b}{2a}$

$d=\frac{16}{2\left(1\right)}$

Cancel the common factor of 16 and 2.

$d=8$

Find the value of e using the formula$e=c-\frac{{b}^{2}}{4a}$

$e=-50$

Substitute the values of$a,d$ and e into the vertex form $a{(x+d)}^{2}+e$

${(x+8)}^{2}-50$

Set y equal to the new right side.

$y={(x+8)}^{2}-50$

Use the form

Consider the vertex form of a parabola.

Substitute the values of a and b into the formula

Cancel the common factor of 16 and 2.

Find the value of e using the formula

Substitute the values of

Set y equal to the new right side.

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