Find the Vertex Form y=-5x^2+x+1

$y = - 5 x^{2} + x + 1$

Step 1

Complete the square for $- 5 x^{2} + x + 1$ .

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

$a = - 5$

$b = 1$

$c = 1$

Consider the vertex form of a parabola.

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

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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

$d = \frac{1}{2 \cdot - 5}$

Simplify the right side.

Multiply $2$ by $- 5$ .

$d = \frac{1}{- 10}$

Move the negative in front of the fraction.

$d = - \frac{1}{10}$

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

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 1 - \frac{1^{2}}{4 \cdot - 5}$

Simplify the right side.

Simplify each term.

One to any power is one.

$e = 1 - \frac{1}{4 \cdot - 5}$

Multiply $4$ by $- 5$ .

$e = 1 - \frac{1}{- 20}$

Move the negative in front of the fraction.

$e = 1 - - \frac{1}{20}$

Multiply $- - \frac{1}{20}$ .

Multiply $- 1$ by $- 1$ .

$e = 1 + 1 (\frac{1}{20})$

Multiply $\frac{1}{20}$ by $1$ .

$e = 1 + \frac{1}{20}$

Write $1$ as a fraction with a common denominator.

$e = \frac{20}{20} + \frac{1}{20}$

Combine the numerators over the common denominator.

$e = \frac{20 + 1}{20}$

Add $20$ and $1$ .

$e = \frac{21}{20}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 5 {(x - \frac{1}{10})}^{2} + \frac{21}{20}$ .

$- 5 {(x - \frac{1}{10})}^{2} + \frac{21}{20}$

Step 2

Set $y$ equal to the new right side.

$y = - 5 {(x - \frac{1}{10})}^{2} + \frac{21}{20}$