# How do you find the intercepts, vertex and graph f(x)=-0.25x^2-3x?

Oct 16, 2016

Vertex is $\left(6 , - 9\right)$
Intercepts are $\left(0 , 0\right)$ and $\left(12 , 0\right)$

#### Explanation:

$f \left(x\right) = 0.25 {x}^{2} - 3 x$
Calculate first derivative $f ' \left(x\right) = 0.5 x - 3$
Let $f ' \left(x\right) = 0$
So $\left(0.5 x - 3 = 0\right) \implies x = \frac{3}{0.5} = 6$
Calculate the second derivative $f ' ' \left(x\right) = 0.5 > 0$
So we have a minimum which is $\left(6 , - 9\right)$
Intercept with the y- axis when $x = 0 \implies y = 0$ So we have the point $\left(0 , 0\right)$
Intercept with the y-axis when $y = 0$
$0.25 {x}^{2} - 3 x = 0 \implies x \left(0.25 x - 3\right) = 0$
So $x = 0$and $x = \frac{3}{0.25} = 12$
So the point is $\left(12 , 0\right)$
Here is a sketch of the graph

graph{0.25x^2-3*x [-20, 20, -10, 10]} /