Find the Vertex F(x)=x^2-4

$F (x) = x^{2} - 4$

Step 1

Rewrite the equation in vertex form.

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

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

$a = 1$

$b = 0$

$c = - 4$

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

$d = \frac{0}{2 \cdot 1}$

Cancel the common factor of $0$ and $2$ .

Factor $2$ out of $0$ .

$d = \frac{2 (0)}{2 \cdot 1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 (0)}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 0}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{0}{1}$

Divide $0$ by $1$ .

$d = 0$

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

$e = - 4 - \frac{0^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$e = - 4 - \frac{0}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = - 4 - \frac{0}{4}$

Divide $0$ by $4$ .

$e = - 4 - 0$

Multiply $- 1$ by $0$ .

$e = - 4 + 0$

Add $- 4$ and $0$ .

$e = - 4$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 0)}^{2} - 4$ .

${(x + 0)}^{2} - 4$

Set $y$ equal to the new right side.

$y = {(x + 0)}^{2} - 4$

Step 2

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

$a = 1$

$h = 0$

$k = - 4$

Step 3

Find the vertex $(h, k)$ .

$(0, - 4)$

Step 4

Find the Vertex F(x)=x^2-4

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