# What is the vertex form of y=−x^2 − 7x + 1 ?

The Vertex Form ${\left(x - - \frac{7}{2}\right)}^{2} = - \left(y - \frac{53}{4}\right)$ with vertex at $\left(- \frac{7}{2} , \frac{53}{4}\right)$

#### Explanation:

We start from the given and do the "Completing the Square Method"

$y = - {x}^{2} - 7 x + 1$

factor out the $- 1$ first

$y = - 1 \cdot \left({x}^{2} + 7 x\right) + 1$

Compute the number to be added and subtracted using the numerical coefficient of x which is the 7. Divide the 7 by 2 and square the result,...that is ${\left(\frac{7}{2}\right)}^{2} = \frac{49}{4}$

$y = - 1 \cdot \left({x}^{2} + 7 x\right) + 1$

$y = - 1 \cdot \left({x}^{2} + 7 x + \frac{49}{4} - \frac{49}{4}\right) + 1$

the first three terms inside the parenthesis forms a PST-perfect square trinomial.

$y = - 1 \cdot \left({x}^{2} + 7 x + \frac{49}{4} - \frac{49}{4}\right) + 1$

$y = - 1 \cdot \left(\left({x}^{2} + 7 x + \frac{49}{4}\right) - \frac{49}{4}\right) + 1$

$y = - 1 \cdot \left({\left(x + \frac{7}{2}\right)}^{2} - \frac{49}{4}\right) + 1$

simplify by multiplying the -1 back and removing the grouping symbol

$y = - 1 {\left(x + \frac{7}{2}\right)}^{2} + \frac{49}{4} + 1$

$y = - 1 {\left(x + \frac{7}{2}\right)}^{2} + \frac{53}{4}$

$y - \frac{53}{4} = - 1 {\left(x + \frac{7}{2}\right)}^{2}$

Let us form the Vertex Form
${\left(x - h\right)}^{2} = \pm 4 p \left(y - k\right)$

${\left(x - - \frac{7}{2}\right)}^{2} = - \left(y - \frac{53}{4}\right)$

Kindly see the graph
graph{(x- -7/2)^2=-(y-53/4)[-30,30,-15,15]}

God bless ....I hope the explanation is useful.