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

Jul 29, 2017

Graph of $f \left(x\right)$ is a parabola with $x -$ intercepts $\left(- 2 , 0\right) \mathmr{and} \left(5 , 0\right)$ and an absolute maximum at $\left(1.5 , 12.25\right)$

#### Explanation:

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

The first two 'important points' are the zeros of $f \left(x\right)$. These occur where $f \left(x\right) = 0$ - I.e. the $x -$intercepts of the function.

To find the zeros: $- \left(x + 2\right) \left(x - 5\right) = 0$

$\therefore x = - 2 \mathmr{and} 5$

Hence the $x -$intercepts are: $\left(- 2 , 0\right) \mathmr{and} \left(5 , 0\right)$

Expanding $f \left(x\right)$

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

$f \left(x\right)$ is a quadratic function of the form $a {x}^{2} + b x + c$. Such a function is represented graphically as a parabola.

The vertex of the parabola occurs at $x = \frac{- b}{2 a}$

i.e where $x = \frac{- 3}{-} 2 = \frac{3}{2} = 1.5$

Since $a < 0$ the vertex will be at the absolute maximum $f \left(x\right)$

$\therefore {f}_{\max} = f \left(\frac{3}{2}\right) = - {\left(\frac{3}{2}\right)}^{2} + 3 \left(\frac{3}{2}\right) + 10$

$= - \frac{9}{4} + \frac{9}{2} + 10 = \frac{9}{4} + 10 = 12.25$

Hence another 'important point' is: ${f}_{\max} = \left(1.5 , 12.25\right)$

We can see these points of the graph of $f \left(x\right)$ below.

graph{-(x+2)(x-5) [-36.52, 36.52, -18.27, 18.27]}