Question

f(x) = \frac{x^2 - 3x - 10}{x + 2}
step 1: find all vertical asymptotes and/or holes and describe the function behavior around them.
step 2: evaluate \lim_{x\to\infty} f(x) and \lim_{x\to-\infty} f(x). write the equations of any horizontal or slant asymptotes, if applicable.
step 3: find all x- and y-intercepts.
step 4: use the work of your group members to graph y = f(x). choose an appropriate scale.

Explanation:

Step 1: Simplify the function

First, factor the numerator: \(x^{2}-3x - 10=(x - 5)(x+2)\). So the function \(f(x)=\frac{(x - 5)(x + 2)}{x + 2}\). We can cancel the common factor \(x + 2\) (for \(x
eq - 2\)), so \(f(x)=x - 5\) (with a hole at \(x=-2\), since \(x=-2\) makes the original denominator zero but the numerator also zero after factoring). To find the \(y\)-coordinate of the hole, substitute \(x = - 2\) into \(f(x)=x - 5\), we get \(y=-2-5=-7\). So there is a hole at \((-2,-7)\) and no vertical asymptote (because the common factor cancels out, removing the vertical asymptote candidate \(x=-2\)).

Step 2: Evaluate the limits at infinity

For \(\lim_{x
ightarrow\infty}f(x)\) and \(\lim_{x
ightarrow-\infty}f(x)\), since \(f(x)=x - 5\) (for \(x
eq - 2\)), the limit as \(x
ightarrow\infty\) of \(x - 5\) is \(\infty\) and as \(x
ightarrow-\infty\) is \(-\infty\). Since the degree of the numerator (after canceling, the simplified function is linear, degree 1) is greater than the degree of the denominator (degree 0, since after canceling the denominator is effectively 1), there is a slant asymptote. The slant asymptote is the simplified linear function \(y=x - 5\) (note that we have to consider the domain exclusion, but for the asymptote, the slant asymptote is \(y=x - 5\) because as \(x\) approaches \(\pm\infty\), the function \(f(x)\) behaves like \(y=x - 5\)).

Step 3: Find intercepts

• x-intercept: Set \(y = 0\) in \(f(x)=x - 5\) (since the hole doesn't affect the intercepts of the main part of the function), so \(x-5=0\), \(x = 5\). So the x-intercept is \((5,0)\).
• y-intercept: Set \(x = 0\) in \(f(x)=x - 5\), we get \(y=0 - 5=-5\). So the y-intercept is \((0,-5)\).

Step 4: Graphing

The function is a line \(y=x - 5\) with a hole at \((-2,-7)\). So we draw the line \(y=x - 5\) and mark a hole (open circle) at \((-2,-7)\).

Answer:

• Step 1: Hole at \((-2, - 7)\), no vertical asymptote.
• Step 2: \(\lim_{x

ightarrow\infty}f(x)=\infty\), \(\lim_{x
ightarrow-\infty}f(x)=-\infty\), slant asymptote \(y = x-5\).

• Step 3: x-intercept \((5,0)\), y-intercept \((0,-5)\).
• Step 4: Graph the line \(y=x - 5\) with an open circle at \((-2,-7)\).