# How do you find the vertex and the intercepts for y = (x+4)(x-6) ?

Aug 18, 2017

Vertex is at $\left(1 , - 25\right)$ , $x$ intercepts are at $\left(- 4 , 0\right) \ast \mathmr{and} \ast \left(6 , 0\right)$ and $y$ intercept is at $\left(0 , - 24\right)$

#### Explanation:

$y = \left(x + 4\right) \left(x - 6\right) \mathmr{and} y = {x}^{2} - 2 x - 24$ or

$y = {x}^{2} - 2 x + 1 - 1 - 24 \mathmr{and} y = {\left(x - 1\right)}^{2} - 25$

Comparing with vertex form of equation y=a(x-h)^2+k ; (h,k)

being vertex , here $h = 1 , k = - 25$ , So vertex is at $\left(1 , - 25\right)$

To find $y$ intercept , putting $x = 0$ in the equation we get

$y = \left(x + 4\right) \left(x - 6\right) \mathmr{and} y = \left(0 + 4\right) \left(0 - 6\right) = - 24$ ,

$y$ intercept is at $y = - 24$ or at $\left(0 , - 24\right)$ . To find $x$ intercept ,

putting $y = 0$ in the equation we get , $0 = \left(x + 4\right) \left(x - 6\right)$

$\therefore x = - 4 , x = 6 \therefore x$ intercepts are at $x = - 4 , x = 6$or at

$\left(- 4 , 0\right) \mathmr{and} \left(6 , 0\right)$.

graph{(x+4)(x-6) [-80, 80, -40, 40]} [Ans]