Test 1 Section 4 #36

You’re in the calculator section, so the fastest way to get this is by graphing. Just type the function in carefully, and graph it:

t1s4-36-1

Those vertical asymptotes around x = 3 tell you that 3 is probably the answer. Want to make sure? Look at your Table:

t1s4-36-3

Yep, the function is undefined at = 3.

Whenever I post a calculator-based solution, though, people always ask me how to solve without a calculator. Of course, there’s a way, but here is a good place for me to recommend to you that if you don’t have one of your own, you find a calculator you can use on test day (maybe you can borrow one from school) and take the time now to get used to using it. Anyway, the math:

    \begin{align*}h(x)&=\dfrac{1}{(x-5)^2+4(x-5)+4}\end{align*}

What they’re trying to show you there with the (x – 5)s is that you might want to consider them a whole value and simplify this by substituting for (x – 5). Let’s say (x – 5) = p for a moment. Then we can say:

    \begin{align*}h(x)&=\dfrac{1}{p^2+4p+4}\end{align*}

Of course, that’s a binomial square, so we can say:

    \begin{align*}h(x)&=\dfrac{1}{(p+2)^2}\end{align*}

Now reintroduce (x – 5) for p:

    \begin{align*}h(x)&=\dfrac{1}{(x-5+2)^2}\end{align*}

    \begin{align*}h(x)&=\dfrac{1}{(x-3)^2}\end{align*}

And that is why the function is undefined at x = 3: the denominator equals zero when x = 3.

Note that you can also simplify to h(x)=\dfrac{1}{(x-3)^2} if you expand the denominator rather than doing the substitution I just did. That’s not too much more labor intensive and is arguably more straightforward. That solution looks like this:

    \begin{align*}h(x)&=\dfrac{1}{(x-5)^2+4(x-5)+4}\end{align*}

    \begin{align*}h(x)&=\dfrac{1}{x^2-10x+25+4x-20+4}\end{align*}

    \begin{align*}h(x)&=\dfrac{1}{x^2-6x+9}\end{align*}

    \begin{align*}h(x)&=\dfrac{1}{(x-3)^2}\end{align*}