Find the Vertex Form f(x)=-3x^2+6x-5

$f (x) = - 3 x^{2} + 6 x - 5$

Step 1

Write $f (x) = - 3 x^{2} + 6 x - 5$ as an equation.

$y = - 3 x^{2} + 6 x - 5$

Step 2

Complete the square for $- 3 x^{2} + 6 x - 5$ .

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

$a = - 3$

$b = 6$

$c = - 5$

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

$d = \frac{6}{2 \cdot - 3}$

Simplify the right side.

Cancel the common factor of $6$ and $2$ .

Factor $2$ out of $6$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot - 3$ .

$d = \frac{2 \cdot 3}{2 (- 3)}$

Cancel the common factor.

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

Rewrite the expression.

$d = \frac{3}{- 3}$

Cancel the common factor of $3$ and $- 3$ .

Factor $3$ out of $3$ .

$d = \frac{3 (1)}{- 3}$

Move the negative one from the denominator of $\frac{1}{- 1}$ .

$d = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$d = - 1$

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

$e = - 5 - \frac{6^{2}}{4 \cdot - 3}$

Simplify the right side.

Simplify each term.

Raise $6$ to the power of $2$ .

$e = - 5 - \frac{36}{4 \cdot - 3}$

Multiply $4$ by $- 3$ .

$e = - 5 - \frac{36}{- 12}$

Divide $36$ by $- 12$ .

$e = - 5 - - 3$

Multiply $- 1$ by $- 3$ .

$e = - 5 + 3$

Add $- 5$ and $3$ .

$e = - 2$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 3 {(x - 1)}^{2} - 2$ .

$- 3 {(x - 1)}^{2} - 2$

Step 3

Set $y$ equal to the new right side.

$y = - 3 {(x - 1)}^{2} - 2$

Find the Vertex Form f(x)=-3x^2+6x-5

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