Find the Vertex Form y=-3x^2-5x+4

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

Step 1

Complete the square for $- 3 x^{2} - 5 x + 4$ .

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

$a = - 3$

$b = - 5$

$c = 4$

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

$d = \frac{- 5}{2 \cdot - 3}$

Simplify the right side.

Multiply $2$ by $- 3$ .

$d = \frac{- 5}{- 6}$

Dividing two negative values results in a positive value.

$d = \frac{5}{6}$

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

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $- 3$ .

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

Move the negative in front of the fraction.

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

Multiply $- - \frac{25}{12}$ .

Multiply $- 1$ by $- 1$ .

$e = 4 + 1 (\frac{25}{12})$

Multiply $\frac{25}{12}$ by $1$ .

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

To write $4$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

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

Combine $4$ and $\frac{12}{12}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $4$ by $12$ .

$e = \frac{48 + 25}{12}$

Add $48$ and $25$ .

$e = \frac{73}{12}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 3 {(x + \frac{5}{6})}^{2} + \frac{73}{12}$ .

$- 3 {(x + \frac{5}{6})}^{2} + \frac{73}{12}$

Step 2

Set $y$ equal to the new right side.

$y = - 3 {(x + \frac{5}{6})}^{2} + \frac{73}{12}$

Find the Vertex Form y=-3x^2-5x+4

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