Graphing Absolute-Value Functions

Inside the absolute-value bars of this function, I've got a quadratic. Without the absolute-value bars, the graph of the quadratic inside the bars is a generic parabola.

I can confirm (by factoring) that the x-intercepts are at x = –1 and x = 4. I can use a formula to confirm that the vertex is at (1.5, –6.25). The y-intercept is at y = –4.

The "regular" parabola's graph looks like this:

Between the two x-intercepts, the graph goes below the x-axis, which means that the output values from the function (that is, the y-values) are negative. But the quadratic is inside absolute-value bars, and those bars will flip that negative part in the middle, redrawing that portion to be above the x-axis. So I should expect the graph to look like this:

So now I have a good idea of what I should expect my absolute-value graph to be. (I did the above "thinking" on scratch-paper; it will not be part of my hand-in answer. But doing this pre-check is very helpful, by reinforcing what I've learned previously.)

Here's my T-chart:

I've got two consecutive x-values with the same y-value. This doesn't mean that there's a horizontal line between (1, 6) and (2,–6); instead, it means that the vertex is halfway between these two x-values, and a little bit above the y-values. I'll want to be careful to draw a nice arc between these two points.

My graph looks like this:

They've given me the absolute value of a cubic. From what I know about the behavior of polynomials, I know that the cubic inside the absolute-value bars comes up from the bottom, maybe flexes a bit in the middle, and then heads upward forever. Since I'll be using the absolute values, I know that the left-hand segment of the graph will be flipped up and over the x-axis.

I can find the x-intercepts by factoring.

I'll need some additional plot points, to be certain of where the graph is going. My T-chart, without the computations, is:

Together with the intercepts that I found, my graph ends up looking like this: