# What is the vertex of  y= 4(x+2)^2-2x^2-3x-1?

Dec 20, 2015

Vertex : $\left(- \frac{13}{4} , - \frac{49}{8}\right)$

Vertex form: $y = 2 {\left(x + \frac{13}{4}\right)}^{2} - \frac{49}{8}$

#### Explanation:

Step 1: Expand/multiply the function so it can be int eh standard form of $y = a {x}^{2} + b c + c$

Given
$y = 4 {\left(x + 2\right)}^{2} - 2 x - 3 x - 1$

$= 4 \left(x + 2\right) \left(x + 2\right) - 2 {x}^{2} - 3 x - 1$

$= 4 \left({x}^{2} + 2 x + 2 x + 4\right) - 2 {x}^{2} - 3 x - 1$

$= 4 \left({x}^{2} + 4 x + 4\right) = 2 {x}^{2} - 3 x - 1$

$= 4 {x}^{2} + 16 x + 16 - 2 {x}^{2} - 3 x - 1$

$= 2 {x}^{2} + 13 x + 15$

$a = 2 , \text{ " " b= 13, " " } c = 15$

The formula for vertex is $\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$

${x}_{v e r t e x} = - \frac{b}{2 a} = h$

${x}_{v e r t e x} = \frac{- 13}{2 \cdot 2} = - \frac{13}{4}$

${y}_{v e r t e x} = f \left(- \frac{b}{2 a}\right) = k$

$f \left(- \frac{13}{4}\right) = 2 {\left(- \frac{13}{4}\right)}^{2} + 13 \left(- \frac{13}{4}\right) + 15$

$= 2 \left(\frac{169}{16}\right) - \frac{169}{4} + 15 = - \frac{49}{8}$

Vertex form: $y = a {\left(x - h\right)}^{2} + k$

$y = 2 {\left(x + \frac{13}{4}\right)}^{2} - \frac{49}{8}$