## Thursday, January 04, 2018

### Problem of the Day: Quadratic and Absolute Value Functions

During the break, an online friend, who knows I'm a math teacher (the user handle @mrburkemath is generally a tip-off), sent me the following math review problem:

"Given the functions h(x) = |x - 4| + 1 and k(x) = x2 + 3, which intervals contain a value of x for which h(x) = k(x)?
This was followed by a list of intervals in the form number < x < number. I've left these out as I have to assume the entire problem is copyrighted from a review book. I'm hoping my excerpt is covered by "fair use".

First thing I said what, in a purely multiple-choice format, just plug the two functions into a calculator (or use an online app, if you're at home without one) and look for the answers. Then select the intervals that include those values.

Otherwise, we can work it out. First thing to realize in that dealing with quadratics and absolute values, there can be, at most, two real number answers. (The problem didn't involve imaginary roots.)

The friend wasn't sure if she was supposed to set each equal to zero and solve, or set them equal to each other. It's in the question: h(x) = k(x).

So here is what we have:

h(x) = k(x)
|x - 4| + 1 = x2 + 3

First thing is to isolate the absolute value, by subtracting 1 from each side:

|x - 4| = x2 + 2

Many of you can skip the mini-review I'm going to do right now.
When solving an absolute value equation like |x - 4| = 7, you have to split the equation into two possibilities, one positive, one negative, and solve each.

 x - 4 = 7 x - 4 = -7

Applying that rule to this equation, you get
When solving an absolute value equation like |x - 4| = 7, you have to split the equation into two possibilities, one positive, one negative, and solve each.

 x - 4 = x2 + 2 x - 4 = -(x2 + 2) x - 4 = x2 + 2 x - 4 = -x2 - 2 0 = x2 - x + 6 x2 + x - 2 = 0 No real roots Factor (x + 2)(x - 1) = 0 x = -2 or x = 1

If you check the discriminant (b2 - 4ac) for the first equation, you will get a number less than 0, meaning that there are no real roots.

Going back to the original problem, they wanted you to select any of the intervals that contained either -2 or 1 or both.

Had they asked for the points of intersection, the values of the functions for those x values, then you have to plug them in:
h(-2) = |-2 - 4| + 1 = |-6| + 1 = 6 + 1 = 7, (-2, 7)
h(1) = |1 - 4| + 1 = |-3| + 1 = 3 + 1 = 4, (1, 4)

Simple, right?