# How do you find the vertex and the intercepts for f(x)=x^2-x-12 ?

May 8, 2018

Vertex is at $\left(0.5 , - 12.25\right)$, x intercepts are at
$\left(- 3 , 0\right) \mathmr{and} \left(4 , 0\right)$ and y intercept is at $\left(0 , - 12\right)$

#### Explanation:

$f \left(x\right) = y = {x}^{2} - x - 12 \mathmr{and} y = \left({x}^{2} - x + \frac{1}{4}\right) - \frac{49}{4}$ or

$y = {\left(x - 0.5\right)}^{2} - 12.25$ Comparing with vertex form of

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

here $h = 0.5 , k = - 12.25 \therefore$ Vertex is at $\left(0.5 , - 12.25\right)$ .

y intercept is found by putting $x = 0$ in the equation

$y = {x}^{2} - x - 12 \therefore y = - 12$ or at $\left(0 , - 12\right)$

y intercept is at $\left(0 , - 12\right)$

x intercept is found by putting $y = 0$ in the equation

$y = {x}^{2} - x - 12 \mathmr{and} {x}^{2} - x - 12 = 0$ or

${x}^{2} - 4 x + 3 x - 12 = 0 \mathmr{and} x \left(x - 4\right) + 3 \left(x - 4\right) = 0$ or

$\left(x - 4\right) \left(x + 3\right) = 0 \therefore x = - 3 \mathmr{and} x = 4$

x intercepts are at $\left(- 3 , 0\right) \mathmr{and} \left(4 , 0\right)$

graph{x^2-x-12 [-40, 40, -20, 20]} [Ans]