r.5 homework asymptotoes and holes identify the points of discontinuity of each. pods 1) f(x)=(x + 4)/(x²…

Explanation:

Step1: Recall the condition for discontinuity

A rational - function $f(x)=\frac{g(x)}{h(x)}$ is discontinuous at values of $x$ for which $h(x) = 0$.

Step2: Solve for discontinuity in 1)

For $f(x)=\frac{x + 4}{x^{2}-4}=\frac{x + 4}{(x + 2)(x - 2)}$, set $(x + 2)(x - 2)=0$. $x+2 = 0$ gives $x=-2$ and $x - 2=0$ gives $x = 2$.

Step3: Solve for discontinuity in 2)

For $f(x)=\frac{-x^{2}+3x}{x^{2}-7x + 12}=\frac{-x(x - 3)}{(x-3)(x - 4)}$, set $(x - 3)(x - 4)=0$. $x-3 = 0$ gives $x = 3$ and $x - 4=0$ gives $x = 4$. But $x = 3$ is a removable discontinuity (hole).

Step4: Solve for discontinuity in 3)

For $f(x)=\frac{2}{x^{2}+2x - 3}=\frac{2}{(x + 3)(x - 1)}$, set $(x + 3)(x - 1)=0$. $x+3 = 0$ gives $x=-3$ and $x - 1=0$ gives $x = 1$.

Step5: Solve for discontinuity in 4)

For $f(x)=\frac{2x + 4}{x^{2}+2x}=\frac{2(x + 2)}{x(x + 2)}$, set $x(x + 2)=0$. $x=0$ and $x=-2$. But $x=-2$ is a removable discontinuity (hole).

Step6: Find holes and asymptotes in 5)

For $f(x)=\frac{x^{2}-2x - 8}{3x + 3}=\frac{(x-4)(x + 2)}{3(x + 1)}$, vertical asymptote at $x=-1$ (since $3(x + 1)=0$ when $x=-1$), no holes.

Step7: Find holes and asymptotes in 6)

For $f(x)=\frac{2x + 6}{x^{2}+4x + 3}=\frac{2(x + 3)}{(x + 1)(x+3)}$, hole at $x=-3$ (because the factor $(x + 3)$ cancels) and vertical asymptote at $x=-1$ (since $x + 1=0$ when $x=-1$).

Answer:

1. Discontinuities at $x=-2,x = 2$.
2. Discontinuities at $x = 3,x = 4$; hole at $x = 3$.
3. Discontinuities at $x=-3,x = 1$.
4. Discontinuities at $x=0,x=-2$; hole at $x=-2$.
5. Vertical asymptote at $x=-1$, no holes.
6. Hole at $x=-3$, vertical asymptote at $x=-1$.