Question

How do you use the important points to sketch the graph of `f(x)=2(x+4)(x-2)`?

Answer 1

Pease see below.

Explanation:

This is the equation of a parabola in intercept form as intercepts on `x`-axis are given by `x+4=0` and `x-2=0` i.e. `-4` and `2`. Hence two points on graph are `(-4,0)` and `(2,0)`.

As coefficient of `x^2` is positive, we have a minima at vertex and the parabola is vertical and axis of symmetry is midway between the two intercepts i.e. at `x=(-4+2)/2=-1` and at `x=-1` `f(x)=-18` and vertex is `(-1,-18)`.

We can find additional points on graph by selecting a few more values of `x` around `-1`, say `[-5,-3,-2,0,1,3}` which give corresponding value of `f(x)` as `{14,-10,-16,-16,-10,14}` and plotting these, we get the graph as shown below.

graph{2(x+4)(x-2) [-40, 40, -20, 20]}