Find the Vertex f(x)=-3x^2+6x-4

$f (x) = - 3 x^{2} + 6 x - 4$

Rewrite the equation in vertex form.

Complete the square for $- 3 x^{2} + 6 x - 4$ .

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

$a = - 3$

$b = 6$

$c = - 4$

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

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

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

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $- 3$ .

$e = - 4 - \frac{36}{- 12}$

Divide $36$ by $- 12$ .

$e = - 4 - - 3$

Multiply $- 1$ by $- 3$ .

$e = - 4 + 3$

Add $- 4$ and $3$ .

$e = - 1$

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

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

Set $y$ equal to the new right side.

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

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = - 3$

$h = 1$

$k = - 1$

Find the vertex $(h, k)$ .

$(1, - 1)$

Find the Vertex f(x)=-3x^2+6x-4

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