Find the Vertex Form y=-x^2+5x-3

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

Step 1

Complete the square for $- x^{2} + 5 x - 3$ .

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

$a = - 1$

$b = 5$

$c = - 3$

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 $d = \frac{b}{2 a}$ .

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

Simplify the right side.

Multiply $2$ by $- 1$ .

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

Move the negative in front of the fraction.

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

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = - 3 - \frac{5^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Raise $5$ to the power of $2$ .

$e = - 3 - \frac{25}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = - 3 - \frac{25}{- 4}$

Move the negative in front of the fraction.

$e = - 3 - - \frac{25}{4}$

Multiply $- - \frac{25}{4}$ .

Multiply $- 1$ by $- 1$ .

$e = - 3 + 1 (\frac{25}{4})$

Multiply $\frac{25}{4}$ by $1$ .

$e = - 3 + \frac{25}{4}$

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$e = - 3 \cdot \frac{4}{4} + \frac{25}{4}$

Combine $- 3$ and $\frac{4}{4}$ .

$e = \frac{- 3 \cdot 4}{4} + \frac{25}{4}$

Combine the numerators over the common denominator.

$e = \frac{- 3 \cdot 4 + 25}{4}$

Simplify the numerator.

Multiply $- 3$ by $4$ .

$e = \frac{- 12 + 25}{4}$

Add $- 12$ and $25$ .

$e = \frac{13}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x - \frac{5}{2})}^{2} + \frac{13}{4}$ .

$- {(x - \frac{5}{2})}^{2} + \frac{13}{4}$

Step 2

Set $y$ equal to the new right side.

$y = - {(x - \frac{5}{2})}^{2} + \frac{13}{4}$

Find the Vertex Form y=-x^2+5x-3

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