Question

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

Answer 1

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

Explanation:

Given:

`h(x) = -2(x-4)(x+2)`

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 `(4, 0)` and `(-2, 0)`.

• 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(color(blue)(1)-4)(color(blue)(1)+2) = -2(-3)(3) = 18`

  So the vertex is at `(1, 18)`

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

  `y = -2(color(blue)(0)-4)(color(blue)(0)+2) = -2(-4)(2) = 16`

  That is `(0, 16)`

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]}