Question

which graph represents the rational function $f(x) = \frac{x^2 - 16}{x^2 - 2x - 8}$?

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 - 16\) as \((x - 4)(x + 4)\) (difference of squares: \(a^2 - b^2=(a - b)(a + b)\), here \(a = x\), \(b = 4\)).
Factor \(x^2 - 2x - 8\): find two numbers that multiply to \(-8\) and add to \(-2\), which are \(-4\) and \(2\). So \(x^2 - 2x - 8=(x - 4)(x + 2)\).
Now the function is \(f(x)=\frac{(x - 4)(x + 4)}{(x - 4)(x + 2)}\), \(x
eq4\) (to avoid division by zero). Cancel \((x - 4)\) (for \(x
eq4\)): \(f(x)=\frac{x + 4}{x + 2}\), \(x
eq4\).

Step2: Simplify the simplified function

Rewrite \(\frac{x + 4}{x + 2}\) as \(\frac{(x + 2)+2}{x + 2}=1+\frac{2}{x + 2}\). This is a transformation of \(y=\frac{2}{x}\): shifted left by 2 units and up by 1 unit.

Step3: Find vertical asymptote

Vertical asymptote where denominator is zero: \(x + 2 = 0\Rightarrow x=-2\).

Step4: Find horizontal asymptote

For rational functions \(\frac{ax + b}{cx + d}\) (degree of numerator = degree of denominator), horizontal asymptote is \(\frac{a}{c}\). Here \(a = 1\), \(c = 1\), so \(y = 1\).

Step5: Find x-intercept

Set \(f(x)=0\): \(\frac{x + 4}{x + 2}=0\Rightarrow x + 4 = 0\Rightarrow x=-4\) (and \(x
eq - 2\), so valid).

Step6: Find y-intercept

Set \(x = 0\): \(f(0)=\frac{0 + 4}{0 + 2}=2\), so \(y\)-intercept at \((0,2)\).

Step7: Analyze the graph

The graph should have a vertical asymptote at \(x=-2\), horizontal asymptote at \(y = 1\), x-intercept at \((-4,0)\), y-intercept at \((0,2)\), and a hole at \(x = 4\) (since we canceled \((x - 4)\), the function is undefined at \(x = 4\) but approaches the value of \(\frac{4 + 4}{4 + 2}=\frac{8}{6}=\frac{4}{3}\) near \(x = 4\)). The given graph (from the image) has a vertical asymptote near \(x=-2\) (since the left and right branches near \(x=-2\) go to \(\pm\infty\)), horizontal asymptote near \(y = 1\) (the right branch approaches \(y = 1\) as \(x\to\pm\infty\)), x-intercept at \(x=-4\) (crosses x-axis at \(x=-4\)), and y-intercept around \(y = 2\) (matches our calculation). Also, the hole at \(x = 4\) is a point not on the simplified line but the original function skips it. The graph shown (with branches near \(x=-2\), crossing x-axis at \(x=-4\), y-intercept at \(y = 2\), and approaching \(y = 1\)) matches these features.

Answer:

The graph shown (with vertical asymptote at \(x=-2\), horizontal asymptote at \(y = 1\), x-intercept at \((-4,0)\), y-intercept at \((0,2)\)) represents the function \(f(x)=\frac{x^2 - 16}{x^2 - 2x - 8}\). (If choosing from the given graph in the image, it's the one with the described features.)