Graphing Rational Functions

Sol 1

1) Since $f$ is a non-constant polynomial, there are no asymptotes for its graph.

2) a) $f(0)=9$, so the y-intercept is 9.

b) $f(x)=x^4-10x^2+9=(x^2-9)(x^2-1)=(x-3)(x+3)(x-1)(x+1)$, so $f(x)=0$ iff $x=\pm3$ or $x=\pm1$ and the x-intercepts are -3,3,-1, and 1.

3) Using the facts that $f(0)=9$ and that all the exponents are odd, we get the following sign chart for $f(x)$:

4) Since $f(-x)=f(x)$, $f$ is an even function and therefore its graph is symmetric around the y-axis.

5) Using the above information, we get the following graph:

Sol 2

1) a) The vertical asymptote is the line $x=2$.

b) Since $P(x)$ and $Q(x)$ have the same degree and they both have leading coefficient 1, the horizontal asymptote is the line $y=1/1$ or $y=1$.

2) a) $f(0)=-1$, so the y-intercept is -1.

b) $f(x)=0\iff x+2=0\iff x=-2$, so the only x-intercept is -2.

3) Using the facts that $f(0)=-1$ and that all the exponents are odd, we get the following sign chart for $f(x)$:

4) a) Setting $f(x)=1$ and solving gives $\frac{x+2}{x-2}=1$, so $x+2=x-2$ and $2=-2$. Therefore there is no solution, so the graph of $f$ does not cross the horizontal asymptote.

b) $f$ is neither even nor odd, so the graph is not symmetric about the y-axis or the origin.

5) Using the information found above, we get the following graph:

Sol 3 [Compare this example to the previous example.]

1) a) Since $f(x)=\frac{x^2+3x+2}{x^2-x-2}=\frac{(x+2)(x+1)}{(x-2)(x+1)}= \frac{x+2}{x-2}$ for $x\ne -1$, the only vertical asymptote is the line $x=2$.

b) Since $P(x)$ and $Q(x)$ have the same degree and they both have leading coefficient 1, the horizontal asymptote is the line $y=1/1$ or $y=1$.

2) a) a) $f(0)=-1$, so the y-intercept is -1.

b) Since $f(x)=\frac{x+2}{x-2}$ for $x\ne -1$, $f(x)=0\iff x+2=0\iff x=-2$; so the only x-intercept is -2.

3) Since $f(0)=-1$ and both exponents are odd (and $f$ is undefined at -1),

we get the following sign chart for $f(x)$:

4) a) Solving $f(x)=1$ gives $x^2+3x+2=x^2-x-2$ or $4x=-4$ or $x=-1$. Since $f$ is undefined at -1, though, its graph does not cross the horizontal asymptote.

b) $f$ is neither even nor odd, so the graph is not symmetric about the y-axis or the origin.

5) Using the above information, we get the graph shown below:

Sol 4

1) a) Since $f(x)=\frac{x-4}{x^2+x-2}=\frac{x-4}{(x+2)(x-1)}$, the vertical asymptotes are the lines $x=-2$ and $x=1$.

b) Since $\deg(P(x))<\deg(Q(x))$, the horizontal asymptote is the line $y=0$ (the x-axis).

2) a) $f(0)=2$, so the y-intercept is 2.

b) $f(x)=0\iff x-4=0\iff x=4$, so the only x-intercept is 4.

3) Since $f(0)=2$ and all the exponents are odd,

we get the following sign chart for $f(x)$:

4) a) Setting $f(x)=0$ and solving gives $x=4$, so the graph intersects the horizontal asymptote at the point $(4,0)$.

b) $f$ is neither even nor odd, so the graph is not symmetric about the y-axis or the origin.

5) Using the above information, we get the graph shown below:

Sol 5

1) a) Since $f(x)=\frac{x^2}{16-x^2}=\frac{x^2}{(4-x)(4+x)}$, the vertical asymptotes are the lines $x=4$ and $x=-4$.

b) Since $P(x)$ and $Q(x)$ have the same degree, the horizontal asymptote is the line $y=1/(-1)$ or $y=-1$.

2) a) $f(0)=0$, so the y-intercept is 0.

b) $f(x)=0 \iff x^2=0 \iff x=0$, so the only x-intercept is 0.

3) Using the facts that $f(1)=1/15>0$ and the sign of $f(x)$ changes at 4 and at -4 but does not change at 0,

we get the following sign chart for $f(x)$:

4) a) Setting $f(x)=-1$ gives $\frac{x^2}{16-x^2}=-1$, so $x^2=-(16-x^2)=-16+x^2$ and $0=-16$. Therefore there is no solution, so the graph of $f$ does not cross its horizontal asymptote.

b) Since $f(-x)=f(x)$, $f$ is an even function and therefore its graph is symmetric about the y-axis.

5) Using the information we have found, we get the following graph:

Sol 6

1) a) $f(x)=x-4+3/x=\frac{x(x-4)+3}{x}=\frac{x^2-4x+3}{x}$, so the only vertical asymptote is the line $x=0$ (the y-axis).

b) Since $\deg(P(x))>\deg(Q(x))$, there is no horizontal asymptote; but since $\deg(P(x))=\deg(Q(x))+1$, there is a slanted asymptote:

Dividing $P(x)$ by $Q(x)$ gives the original equation $f(x)=x-4+3/x$,

so the line $y=x-4$ is the slanted asymptote.

2) a) Since $f(0)$ is undefined, there is no y-intercept.

b) $f(x)=0 \iff x^2-4x+3=0 \iff (x-3)(x-1)=0 \iff x=3$ or $x=1$, so the x-intercepts are 1 and 3.

3) Since $f(x)=\frac{(x-3)(x-1)}{x}$, using the facts that $f(2)=-1/2<0$ and all the exponents are odd gives the following sign chart for $f(x)$:

4) a) Setting $f(x)=x-4$ and solving gives $x-4+3/x=x-4$ or $3/x=0$ or $3=0$. Therefore there is no solution, so the graph of $f$ does not intersect its slanted asymptote.

b) $f$ is neither even nor odd, so the graph is not symmetric about the y-axis or the origin.

5) Using the above information, we get the graph shown below:

Sol 7

1) a) Since $f(x)=x+\frac{1}{x-2}=\frac{x(x-2)+1}{x-2}=\frac{x^2-2x+1}{x-2}$,

the line $x=2$ is the only vertical asymptote.

b) Since $\deg(P(x))>\deg(Q(x))$, there is no horizontal asymptote; but since $\deg(P(x))=\deg(Q(x))+1$, there is a slanted asymptote:

Dividing $P(x)$ by $Q(x)$ gives the original equation $f(x)=x+\frac{1}{x-2}$,

so the line $y=x$ is the slanted asymptote.

2) a) $f(0)=-1/2$, so the y-intercept is -1/2.

b) $f(x)=\frac{x^2-2x+1}{x-2}=\frac{(x-1)^2}{x-2}$, so $f(x)=0 \iff (x-1)^2=0 \iff x-1=0 \iff x=1$; and therefore the only x-intercept is 1.

3) Using the facts that $f(0)=-1/2$ and the sign of $f(x)$ changes at 2 but does not change at 1, we get the following sign chart for $f(x)$:

4) a) Setting $f(x)=x$ and solving gives $x+\frac{1}{x-2}=x$, so $\frac{1}{x-2}=0$ and so $1=0$. Therefore there is no solution, so the graph of $f$ does not intersect the slanted asymptote.

b) $f$ is neither even nor odd, so the graph is not symmetric about the y-axis or the origin.

5) Using the above information, we get the graph shown below:

Sol 8

1) a) $f(x)=\frac{x^2-x-6}{x-2}=\frac{(x-3)(x+2)}{x-2}$, so the only vertical asymptote is the line $x=2$.

b) Since $\deg(P(x))>\deg(Q(x))$, there is no horizontal asymptote; but since $\deg(P(x))=\deg(Q(x))+1$, there is a slanted asymptote:

Dividing $P(x)$ by $Q(x)$ gives $\frac{x^2-x-6}{x-2}=x+1+\frac{-4}{x-2}$, so the line $y=x+1$ is the slanted asymptote.

2) a) $f(0)=3$, so the y-intercept is 3.

b) $f(x)=0\iff (x-3)(x+2)=0\iff x=3$ or $x=-2$, so the x-intercepts are -2 and 3.

3) Using the facts that $f(0)=3$ and all the exponents are odd,

we get the following sign chart for $f(x)$:

4) a) Setting $f(x)=x+1$ gives $x+1+\frac{-4}{x-2}=x+1$, so $\frac{-4}{x-2}=0$ and so $-4=0$. Therefore there is no solution, so the graph of $f$ does not intersect the slanted asymptote.

b) $f$ is neither even nor odd, so the graph is not symmetric about the y-axis or the origin.

5) Using the above information, we get the graph shown below:

Sol 9

1) a) Since $f(x)=\frac{x^2-1}{x^2-4x+3}=\frac{(x-1)(x+1)}{(x-1)(x-3)}= \frac{x+1}{x-3}$ for $x\ne 1$, the only vertical asymptote is the line $x=3$.

b) Since $P(x)$ and $Q(x)$ have the same degree, the horizontal asymptote is given by $y=1/1$ or $y=1$.

2) a) $f(0)=-1/3$, so the y-intercept is -1/3.

b) $f(x)=0\iff x+1=0\iff x=-1$, so the only x-intercept is -1.

3) Since $f(0)=-1/3$ and both exponents are odd (and $f$ is undefined at 1),

we get the following sign chart for $f(x)$:

4) a) Setting $f(x)=1$ gives $x^2-1=x^2-4x+3$, so $4x=4$ and $x=1$. However, $f$ is undefined at 1, so the graph of $f$ does not intersect the horizontal asymptote.

b) $f$ is neither even nor odd, so the graph is not symmetric about the y-axis or the origin.

5) From the information above, we get the following graph:

Sol 10

1) a) Since $f(x)=\frac{x^2-2x-8}{x^2-x-2}=\frac{(x-4)(x+2)}{(x-2)(x+1)}$, the vertical asymptotes are the lines $x=2$ and $x=-1$.

b) Since $P(x)$ and $Q(x)$ have the same degree, the horizontal asymptote is the line $y=1/1$ or $y=1$.

2) a) $f(0)=4$, so the y-intercept is 4.

b) $f(x)=0\iff (x-4)(x+2)=0\iff x=4$ or $x=-2$, so the x-intercepts are 4 and -2.

3) Using the facts that $f(0)=4$ and that all the exponents are odd,

we get the following sign chart for $f(x)$:

4) a) Setting $f(x)=1$ gives $\frac{x^2-2x-8}{x^2-x-2}=1$, so $x^2-2x-8=x^2-x-2$ gives $-6=x$ or $x=-6$. Therefore the graph of $f$ crosses the horizontal asymptote at the point $(-6,1)$.

b) $f$ is neither even nor odd, so the graph is not symmetric about the y-axis or the origin.

5) From the information above, we get the following graph:

Sol 11

1) a) $f(x)=\frac{x^3-8}{x^2-3x-4}=\frac{(x-2)(x^2+2x+4)}{(x-4)(x+1)}$, so the vertical asymptotes are $x=4$ and $x=-1$.

b) Since $\deg(P(x))>\deg(Q(x))$, there is no horizontal asymptote; but since $\deg(P(x))=\deg(Q(x))+1$, there is a slanted asymptote:

Dividing $P(x)$ by $Q(x)$ gives $\frac{x^3-8}{x^2-3x-4}=x+3+\frac{13x+4}{x^2-3x-4}$,

so the line $y=x+3$ is the slanted asymptote.

2) a) $f(0)=2$, so the y-intercept is 2.

b) $f(x)=0\iff x^3-8=0\iff x^3=8\iff x=2$, so the only x-intercept is 2.

3) Using the facts that $f(0)=2$ and that the sign of $f(x)$ changes at -1, 2, and 4,

we get the following sign chart for $f(x)$:

4) a) Setting $f(x)=x+3$ gives $x+3+\frac{13x+4}{x^2-3x-4}=x+3$, so $\frac{13x+4}{x^2-3x-4}=0$ and therefore $13x+4=0$ so $x=-4/13$. Since $-4/13+3=35/13$, the graph of $f$ intersects the slanted asymptote at the point $(-4/13,35/13)$.

b) $f$ is neither even nor odd, so the graph is not symmetric about the y-axis or the origin.

5) From the information above, we get the following graph:

Sol 12

1) a) Since $f(x)=\frac{2x^2+4x-6}{x^2+3x-10}=\frac{2(x^2+2x-3)}{x^2+3x-10}= \frac{2(x-1)(x+3)}{(x-2)(x+5)}$, the vertical asymptotes are the lines $x=2$ and $x=-5$.

b) Since $P(x)$ and $Q(x)$ have the same degree, the horizontal asymptote is given by $y=2/1$ or $y=2$.

2) a) $f(0)=3/5$, so the y-intercept is 3/5.

b) $f(x)=0\iff 2(x-1)(x+3)=0\iff x=1$ or $x=-3$, so the x-intercepts are -3 and 1.

3) Since $f(0)=3/5$ and all the exponents are odd,

we get the following sign chart for $f(x)$:

4) a) Setting $f(x)=2$ gives $\frac{2x^2+4x-6}{x^2+3x-10}=2$, so $2x^2+4x-6=2(x^2+3x-10)=2x^2+6x-20$ and $14=2x$ or $x=7$.

Therefore the graph of $f$ intersects the horizontal asymptote at the point $(7,2)$.

b) $f$ is neither even nor odd, so the graph is not symmetric about the y-axis or the origin.

5) From the information above, we get the following graph:

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