Find the Vertex 8x^2+16x-3

$8 x^{2} + 16 x - 3$

Step 1

Rewrite the equation in vertex form.

Complete the square for $8 x^{2} + 16 x - 3$ .

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

$a = 8$

$b = 16$

$c = - 3$

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

$d = \frac{16}{2 \cdot 8}$

Simplify the right side.

Cancel the common factor of $16$ and $2$ .

Factor $2$ out of $16$ .

$d = \frac{2 \cdot 8}{2 \cdot 8}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 8$ .

$d = \frac{2 \cdot 8}{2 (8)}$

Cancel the common factor.

$d = \frac{2 \cdot 8}{2 \cdot 8}$

Rewrite the expression.

$d = \frac{8}{8}$

Cancel the common factor of $8$ .

Cancel the common factor.

$d = \frac{8}{8}$

Rewrite the expression.

$d = 1$

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

$e = - 3 - \frac{16^{2}}{4 \cdot 8}$

Simplify the right side.

Simplify each term.

Raise $16$ to the power of $2$ .

$e = - 3 - \frac{256}{4 \cdot 8}$

Multiply $4$ by $8$ .

$e = - 3 - \frac{256}{32}$

Divide $256$ by $32$ .

$e = - 3 - 1 \cdot 8$

Multiply $- 1$ by $8$ .

$e = - 3 - 8$

Subtract $8$ from $- 3$ .

$e = - 11$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $8 {(x + 1)}^{2} - 11$ .

$8 {(x + 1)}^{2} - 11$

Set $y$ equal to the new right side.

$y = 8 {(x + 1)}^{2} - 11$

Step 2

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

$a = 8$

$h = - 1$

$k = - 11$

Step 3

Find the vertex $(h, k)$ .

$(- 1, - 11)$

Step 4

Find the Vertex 8x^2+16x-3

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