# What are the important points needed to graph f(x) = (x-2)(x + 5)?

Jun 16, 2018

x-intercepts
$x = - 5 , x = 2$
y-intercept
$y = - 10$
vertex: $\left(- \frac{3}{2} , - \frac{49}{4}\right)$

#### Explanation:

You are given the x-intercepts

$\left(x - 2\right) \left(x + 5\right)$
$x = 2$
$x = - 5$

First find y-intercept by multiplying out to standard form $A {x}^{2} + B x + C$ and set x to 0

$f \left(x\right) = \left(x - 2\right) \left(x + 5\right) = {x}^{2} + 3 x - 10$
$f \left(x\right) = {\left(0\right)}^{2} + 3 \left(0\right) - 10 = - 10$

y-intercept is at $y = - 10$

Next convert to vertex form by completing the square

${x}^{2} + 3 x = 10$

Divide coefficient by 2 and square

${\left(\frac{3}{2}\right)}^{2} = \frac{9}{4}$
$\left({x}^{2} + 3 x + \frac{9}{4}\right) = 10 + \frac{9}{4}$

Rewrite

${\left(x + \frac{3}{2}\right)}^{2} = \frac{40}{4} + \frac{9}{4} = \frac{49}{4}$
$f \left(x\right) = {\left(x + \frac{3}{2}\right)}^{2} - \frac{49}{4}$

Vertex is $\left(- \frac{3}{2} , - \frac{49}{4}\right)$ or $\left(- 1.5 , - 12.25\right)$

graph{(x+3/2)^2-49/4 [-21.67, 18.33, -14.08, 5.92]}