Question

at which values of x does the graph of $f(x) = \frac{x^2 - x - 2}{(x^2 - 1)(x^2 - 16)}$ have a vertical asymptote? check all that apply.
-4
-1
1
2
4

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(x^2 - x - 2\). We need two numbers that multiply to \(-2\) and add to \(-1\). Those numbers are \(-2\) and \(1\), so \(x^2 - x - 2=(x - 2)(x + 1)\).

Next, factor the denominator \((x^2 - 1)(x^2 - 16)\). Using the difference of squares formula \(a^2 - b^2=(a - b)(a + b)\), we get:
\(x^2 - 1=(x - 1)(x + 1)\)
\(x^2 - 16=(x - 4)(x + 4)\)
So the denominator is \((x - 1)(x + 1)(x - 4)(x + 4)\).

The function becomes \(f(x)=\frac{(x - 2)(x + 1)}{(x - 1)(x + 1)(x - 4)(x + 4)}\).

Step2: Simplify the function

We can cancel out the common factor \((x + 1)\) (as long as \(x
eq - 1\), because if \(x=-1\), the original function is undefined, but we are looking for vertical asymptotes, not holes). After canceling, we have \(f(x)=\frac{x - 2}{(x - 1)(x - 4)(x + 4)}\) (with the restriction \(x
eq - 1\)).

Step3: Find vertical asymptotes

Vertical asymptotes occur where the denominator of the simplified function is zero (and the numerator is not zero at those points).

Set the denominator \((x - 1)(x - 4)(x + 4)=0\).

• For \(x - 1 = 0\), we get \(x = 1\).
• For \(x - 4 = 0\), we get \(x = 4\).
• For \(x + 4 = 0\), we get \(x=-4\).

We also need to check the point \(x = - 1\). At \(x=-1\), the original function has a common factor in the numerator and denominator, so it is a hole, not a vertical asymptote. At \(x = 2\), the numerator is zero (when \(x = 2\), numerator \((2 - 2)=0\)) and the denominator (after canceling \((x + 1)\)) is \((2 - 1)(2 - 4)(2 + 4)=(1)(-2)(6)=-12
eq0\), so \(x = 2\) is a root (x - intercept), not a vertical asymptote.

Answer:

-4, -1 (wait, no, correction: -4, 1, 4. Wait, let's re - check. Wait, when we canceled \((x + 1)\), the hole is at \(x=-1\), so vertical asymptotes are where the simplified denominator is zero: \(x=-4\), \(x = 1\), \(x = 4\). Wait, in the options, the options are -4, -1, 1, 2, 4.

So from the options:

• For \(x=-4\): denominator of simplified function is zero, so vertical asymptote.
• For \(x=-1\): hole, not vertical asymptote.
• For \(x = 1\): denominator of simplified function is zero, so vertical asymptote.
• For \(x = 2\): numerator is zero, not vertical asymptote.
• For \(x = 4\): denominator of simplified function is zero, so vertical asymptote.

So the values are -4, 1, 4. But let's check the options again. The options are:

• \(-4\)
• \(-1\)
• \(1\)
• \(2\)
• \(4\)

So the correct ones are \(-4\), \(1\), \(4\). But let's confirm with the original function. The vertical asymptote occurs where the denominator is zero and the numerator is not zero (after canceling common factors).

Original denominator: \((x^2 - 1)(x^2 - 16)=(x - 1)(x + 1)(x - 4)(x + 4)\)

Original numerator: \(x^2 - x - 2=(x - 2)(x + 1)\)

So the values that make the denominator zero are \(x = 1\), \(x=-1\), \(x = 4\), \(x=-4\). But at \(x=-1\), the numerator is also zero (since \((-1 - 2)(-1 + 1)=(-3)(0)=0\)), so \(x=-1\) is a hole (removable discontinuity), not a vertical asymptote. At \(x = 1\), numerator is \((1 - 2)(1 + 1)=(-1)(2)=-2
eq0\), so vertical asymptote. At \(x=-4\), numerator is \((-4 - 2)(-4 + 1)=(-6)(-3)=18
eq0\), so vertical asymptote. At \(x = 4\), numerator is \((4 - 2)(4 + 1)=(2)(5)=10
eq0\), so vertical asymptote. At \(x = 2\), denominator is \((4 - 1)(4 + 1)(4 - 4)(4 + 4)=(3)(5)(0)(8)=0\) and numerator is \(4 - 4 - 2=-2
eq0\)? Wait, no, wait \(x = 2\): numerator is \(2^2 - 2 - 2=4 - 2 - 2=0\), denominator is \((4 - 1)(4 + 1)(4 - 4)(4 + 4)=(3)(5)(0)(8)=0\). Wait, I made a mistake earlier. Let's recalculate numerator at \(x = 2\): \(2^2-2 - 2=0\), denominator at \(x = 2\): \((4 - 1)(4 + 1)(4 - 4)(4 + 4)=3\times5\times0\times8 = 0\). So \(x = 2\) is a hole? No, wait the numerator is zero and denominator is zero at \(x = 2\)? Wait no, numerator at \(x = 2\): \(2^2-2 - 2=0\), denominator at \(x = 2\): \((4 - 1)(4 + 1)(4 - 4)(4 + 4)=3\times5\times0\times8 = 0\). Wait, no, the denominator is \((x^2 - 1)(x^2 - 16)\), so at \(x = 2\), \(x^2-1=3\), \(x^2 - 16=-12\), so denominator is \(3\times(-12)=-36
eq0\). Oh! I made a mistake in factoring the denominator earlier. Wait, the denominator is \((x^2 - 1)(x^2 - 16)\), which is \((x - 1)(x + 1)(x - 4)(x + 4)\). At \(x = 2\), \(x - 1=1\), \(x + 1=3\), \(x - 4=-2\), \(x + 4=6\), so denominator is \(1\times3\times(-2)\times6=-36
eq0\), and numerator is \(0\), so \(x = 2\) is an x - intercept. My earlier mistake was in the simplified function. Let's redo the simplification:

\(f(x)=\frac{(x - 2)(x + 1)}{(x - 1)(x + 1)(x - 4)(x + 4)}\), so when we cancel \((x + 1)\) (for \(x
eq - 1\)), we get \(f(x)=\frac{x - 2}{(x - 1)(x - 4)(x + 4)}\) (with \(x
eq - 1\)). So at \(x = 2\), the simplified function has numerator \(0\) and denominator \((2 - 1)(2 - 4)(2 + 4)=1\times(-2)\times6=-12
eq0\), so \(x = 2\) is an x - intercept. At \(x=-1\), original function: numerator is \((-1 - 2)(-1 + 1)=(-3)(0)=0\), denominator is \((1 - 1)(1 + 1)(1 - 4)(1 + 4)=(0)(2)(-3)(5)=0\), so it's a hole. At \(x = 1\): simplified denominator is \((1 - 1)(1 - 4)(1 + 4)=(0)(-3)(5)=0\), numerator is \(1 - 2=-1
eq0\), so vertical asymptote. At \(x=-4\): simplified denominator is \((-4 - 1)(-4 - 4)(-4 + 4)=(-5)(-8)(0)=0\)? No, wait simplified denominator is \((x - 1)(x - 4)(x + 4)\), so at \(x=-4\), \((-4 - 1)(-4 - 4)(-4 + 4)=(-5)(-8)(0)=0\)? No, \((-4 + 4)=0\), so denominator is zero, numerator is \(-4 - 2=-6
eq0\), so vertical asymptote. At \(x = 4\): simplified denominator is \((4 - 1)(4 - 4)(4 + 4)=(3)(0)(8)=0\), numerator is \(4 - 2=2
eq0\), so vertical asymptote. At \(x = 1\): \((1 - 1)=0\), numerator is \(1 - 2=-1
eq0\), so vertical asymptote. At \(x=-4\): \((-4 + 4)=0\), numerator is \(-6
eq0\), vertical asymptote. At \(x = 4\): \((4 - 4)=0\), numerator is \(2
eq0\), vertical asymptote. At \(x=-1\): hole. At \(x = 2\): x - intercept.

So the vertical asymptotes are at \(x=-4\), \(x = 1\), \(x = 4\). So from the options, the correct ones are \(-4\), \(1\), \(4\).