# How do you find the vertex and intercepts for y=2x^2 +11x-6?

Sep 14, 2017

Vertex is at $\left(- 2.75 , - 21.125\right)$ x intercepts are at
$\left(- 6 , 0\right) \mathmr{and} \left(0.5 , 0\right)$, y intercept is at $\left(0 , - 6\right)$

#### Explanation:

$y = 2 {x}^{2} + 11 x - 6 \mathmr{and} y = 2 \left({x}^{2} + \frac{11}{2} x\right) - 6$

Adding $\frac{121}{8}$ on both sides we get ,

$y = 2 \left\{{x}^{2} + \frac{11}{2} x + {\left(\frac{11}{4}\right)}^{2}\right] - \frac{121}{8} - 6$ or

$y = 2 {\left(x + \frac{11}{4}\right)}^{2} - \frac{169}{8}$ or

$y = 2 {\left(x + 2.75\right)}^{2} - 21.125$ . Comparing with standard vertex form

of equation y=a(x-h)^2+k ; (h,k) being vertex , we find here

$h = - 2.75 , k = - 21.125 , a = 2$. So vertex is at $\left(h , k\right)$or

$\left(- 2.75 , - 21.125\right)$ , y intercept can be found by putting

$x = 0$ in the equation as $y = 2 \cdot 0 + 11 \cdot 0 - 6$ or

$y = - 6 \mathmr{and} \left(0 , - 6\right)$ , x intercept can be found by putting $y = 0$

in the equation as $0 = 2 {x}^{2} + 11 x - 6 \mathmr{and} 2 {x}^{2} + 12 x - x - 6 = 0$

or $2 x \left(x + 6\right) - 1 \left(x + 6\right) = 0 \mathmr{and} \left(2 x - 1\right) \left(x + 6\right) = 0$

$\therefore x = 0.5 \mathmr{and} x = - 6$, x intercepts are at $\left(- 6 , 0\right) \mathmr{and} \left(0.5 , 0\right)$

graph{2x^2+11x-6 [-80, 80, -40, 40]} [Ans]