# How do you find the vertex and intercepts for y=2x^2-16x+27?

May 7, 2016

$y = 2 {\left(x - 4\right)}^{2} - 5 \text{ }$ Vertex is at $\left(4 , - 5\right)$
$y -$intercept is at 27

#### Explanation:

$y = 2 {x}^{2} - 16 x + 27$
$y = 2 \left({x}^{2} - 8 x + \frac{27}{2}\right) \text{ }$now complete the square

$y = 2 \left[\textcolor{b l u e}{{x}^{2} - 8 x + 16} - 16 + \frac{27}{2}\right] \text{ } \Rightarrow$ +16-16 = 0

$y = 2 \left[\textcolor{b l u e}{{\left(x - 4\right)}^{2}} - 16 + \frac{27}{2}\right] \text{ }$ $- 16 + \frac{27}{2} = - \frac{5}{2}$

$y = 2 {\left(x - 4\right)}^{2} - 2 \times \frac{5}{2}$

$y = 2 {\left(x - 4\right)}^{2} - 5$