CPM Homework Help : APCALC Problem 1-90

Compute without a calculator Without a calculator, sketch each graph, showing roots, holes, and asymptotes. Then, state the domain in parts (a) and (b) using interval notation and the domain in parts (c) and (d) using set notation.

$y =\large\frac { x ^ { 2 } - 4 } { x ^ { 3 } + 3 x ^ { 2 } - 10 x }$
$y =\large\frac { 9 - x ^ { 2 } } { 2 x + 6 }$
$y =\large\frac { x ^ { 2 } - 9 x - 18 } { x ^ { 2 } + 3 x - 18 }$
$y =\large\frac { x ^ { 2 } - 6 x + 9 } { 15 - 5 x }$

Hint 1(a):

$\text{To make graphing easier, factor first: }y=\frac{(x+2)(x-2)}{x(x+5)(x-2)}$

$\text{Before you simplify, notice all restrictions: }x\neq-5, 0, 2$

Hint 2(a):

$=\frac{(x+2)}{x(x+5)}$

Since $\left(x − 2\right)$ cancelled out, $x = 2$ is the location of a hole.

Now simplify. And notice what was cancelled out.

Hint 3(a):

The $y$ -value of the hole can be found by evaluating the simplified function at $x = 2$ .

$\frac{(2+2)}{(2)(2+5)}=\frac{4}{14}=\frac{2}{7}$

$\text{So the hole has a coordinate point of }\left (-2,\frac{2}{7} \right )$

Hint 4(a):

$x(x + 5)$ did not cancel out, so $x = 0$ and $x = −5$ are vertical asymptotes.

Hint 5(a):

Now find the end behavior. You can long divide or use an approach statement: $x → − ∞$ , $y → 0$ and as $x → ∞$ , $y → 0$ . There is a horizontal asymptote at $y = 0$ .

Hint 6(a):

Last but not least, investigate what is happening in the middle, by exploring the holes and vertical asymptotes. Use the simplified equation:

$y=\frac{(x+2)}{x(x+5)}\text{ and evaluate points near }x=-5, x=0, \text{ and } x=2.$

$x → −5^−, y → −∞$

$x → 0^+, y → +∞$

$x → −5^+, y → +∞$

$x\rightarrow 2^{-},\ y\rightarrow \frac{2}{7}$

$x → 0^−, y → −∞$

$x\rightarrow 2^{+},\ y\rightarrow \frac{2}{7}$

Use the eTool below to view the steps for graphing the first equation.
Click the link at right for the full version of the eTool: Calc 1-90 HW eTool