# How do you graph h(x) = -2(x-4)(x+2) ?

Jan 16, 2017

Identify the intercepts, vertex and axis from the formula...

#### Explanation:

Given:

$h \left(x\right) = - 2 \left(x - 4\right) \left(x + 2\right)$

We can see several properties of the curve from this formula:

• The multiplier of the leading (${x}^{2}$) term is $- 2$, which will result in an inverted parabola with vertex pointing upwards.

• The two $x$ intercepts are at $x = 4$ and $x = - 2$, that is $\left(4 , 0\right)$ and $\left(- 2 , 0\right)$.

• Since a parabola is symmetric about its axis, its axis will be midway between these two $x$ intercepts, at $x = 1$.

• The vertex lies at the intersection of the axis with the parabola, so we can find it by substituting $x = 1$ into the formula:

$y = - 2 \left(\textcolor{b l u e}{1} - 4\right) \left(\textcolor{b l u e}{1} + 2\right) = - 2 \left(- 3\right) \left(3\right) = 18$

So the vertex is at $\left(1 , 18\right)$

• We can find the $y$ intercept by substituting $x = 0$ to find:

$y = - 2 \left(\textcolor{b l u e}{0} - 4\right) \left(\textcolor{b l u e}{0} + 2\right) = - 2 \left(- 4\right) \left(2\right) = 16$

That is $\left(0 , 16\right)$

Here's a graph of the parabola with the features we found indicated:
graph{(y+2(x-4)(x+2))(12(x-4)^2+y^2-0.04)(12(x+2)^2+y^2-0.04)(12(x-1)^2+(y-18)^2-0.04)(x-1)(12x^2+(y-16)^2-0.04) = 0 [-4, 6, -5, 21]}