Rational Functions and Their Graphs: Algebra 2 Math Worksheet Answer Key – PDF Download

Algebra 2 Rational Worksheet Answer Key: problems 1–5

Problem 1: Domain of a rational function

1. \(x+2=0\) at \(x=-2\). Domain: \(x\neq -2\).
2. \(x-7=0\) at \(x=7\). Domain: \(x\neq 7\).
3. \(2x-6=0\) gives \(x=3\). Domain: \(x\neq 3\).
4. \(x^{2}-25=(x-5)(x+5)\), zero at \(x=5\) and \(x=-5\). Domain: \(x\neq 5\text{ and }x\neq -5\).

Problem 2: Evaluate a rational function

\(f(-1)=\frac{0}{-3}=0\); \(f(0)=\frac{1}{-2}=-\frac{1}{2}\); \(f(1)=\frac{2}{-1}=-2\); \(f(3)=\frac{4}{1}=4\); \(f(5)=\frac{6}{3}=2\).

Answers: \(0,\ -\frac{1}{2},\ -2,\ 4,\ 2\).

Problem 3: Table for y = 6/x

Divide 6 by each \(x\).

The points are \((-6,-1),(-3,-2),(-2,-3),(-1,-6),(1,6),(2,3),(3,2),(6,1)\).

Problem 4: Vertical asymptotes

1. \(x+3=0\): \(x=-3\).
2. \(x-1=0\): \(x=1\).
3. \(x^{2}-4=(x-2)(x+2)\): \(x=2\text{ and }x=-2\).
4. \(x^{2}+3x=x(x+3)\). Nothing cancels with \(x+5\), so \(x=0\text{ and }x=-3\).

Problem 5: Horizontal asymptotes

1. Equal degrees: \(\frac{3}{1}\), so \(y=3\).
2. The numerator has the smaller degree: \(y=0\).
3. Equal degrees: \(y=\frac{2}{3}\).
4. The numerator has the larger degree: none.
5. Equal degrees: \(\frac{-4}{2}=-2\), so \(y=-2\).

Algebra 2 Rational Worksheet Answer Key: problems 6–10

Problem 6: Intercepts

1. The numerator is zero when \(2x=6\), so the x-intercept is \((3,0)\). \(f(0)=\frac{-6}{1}=-6\), so the y-intercept is \((0,-6)\).
2. \(x+4=0\) gives \((-4,0)\). \(g(0)=\frac{4}{-2}=-2\), so \((0,-2)\).

Problem 7: Find the hole

1. \(\frac{(x-3)(x+3)}{x-3}=x+3\). At \(x=3\) the value is 6: hole \((3,6)\).
2. \(x^{2}+x-6=(x+3)(x-2)\), so \(g(x)=x+3\) for \(x\neq 2\). Hole \((2,5)\).
3. Cancel \(x-4\) to get \(\frac{x+1}{x+2}\). At \(x=4\): \(\frac{5}{6}\). Hole \((4,\frac{5}{6})\); the factor \(x+2\) gives the vertical asymptote \(x=-2\).

Problem 8: Hole, asymptote and end behavior

Factor: \(f(x)=\frac{(x-1)(x-3)}{(x-1)(x+1)}\). The factor \(x-1\) cancels, leaving \(\frac{x-3}{x+1}\). At \(x=1\) the value is \(\frac{-2}{2}=-1\), so the hole is \((1,-1)\). The factor \(x+1\) stays: vertical asymptote \(x=-1\). Equal degrees with leading coefficients 1 and 1: horizontal asymptote \(y=1\).

Problem 9: Shifts of y = 1/x

1. Left 3, down 2. Asymptotes \(x=-3\) and \(y=-2\).
2. Reflected over the x-axis, right 5, up 4. Asymptotes \(x=5\) and \(y=4\).
3. Up 6 only. Asymptotes \(x=0\) and \(y=6\).

Problem 10: Read a hyperbola

1. The branches close in on \(x=2\) and \(y=1\) (the center is \((2,1)\)).
2. Domain: \(x\neq 2\). Range: \(y\neq 1\).
3. Use \(h=2\), \(k=1\) and the point \((3,4)\): \(4=\frac{a}{1}+1\), so \(a=3\). Equation: \(y=\frac{3}{x-2}+1\). Check \((5,2)\): \(\frac{3}{3}+1=2\).

Algebra 2 Rational Worksheet Answer Key: problems 11–15

Problem 11: Graph a rational function

The asymptotes are \(x=-1\) and \(y=-3\). Points: \(x=0\) gives \(2-3=-1\); \(x=1\) gives \(1-3=-2\); \(x=-2\) gives \(-2-3=-5\); \(x=-3\) gives \(-1-3=-4\).

Problem 12: Equation from a graph

The dashed lines give \(h=-1\) and \(k=2\), so \(y=\frac{a}{x+1}+2\). The point \((0,5)\) gives \(5=\frac{a}{1}+2\), so \(a=3\). Equation: \(y=\frac{3}{x+1}+2\). Check \((2,3)\): \(\frac{3}{3}+2=3\).

Problem 13: A graph with a hole

1. \(\frac{(x-1)(x+1)}{x-1}=x+1\), so \(f(x)=x+1\) for \(x\neq 1\).
2. At \(x=1\) the line has height 2, so the hole is \((1,2)\).
3. Domain: \(x\neq 1\).

Problem 14: Write a rational function

\(h=2\) and \(k=-1\). Substitute \((3,1)\): \(1=\frac{a}{3-2}-1\), so \(a=2\). Function: \(y=\frac{2}{x-2}-1\).

Problem 15: Inverse variation constant

1. \(k=xy=5\cdot 8=40\), so \(y=\frac{40}{x}\).
2. \(y=\frac{40}{20}=2\). Answer: \(y=2\).
3. \(16=\frac{40}{x}\) gives \(x=\frac{40}{16}=\frac{5}{2}\). Answer: \(x=2.5\).

Algebra 2 Rational Worksheet Answer Key: problems 16–20

Problem 16: Is it inverse variation?

1. Products: \(36,36,36,36\). Yes, \(k=36\).
2. Products: \(5,20,45,80\) are not constant. The ratio \(\frac{y}{x}=5\) is constant, so this is direct variation: not inverse.
3. Products: \(60,60,60,60\). Yes, \(k=60\).

Problem 17: Inverse variation from a graph

1. \(k=2\cdot 12=24\) (check \(4\cdot 6=24\)). Equation: \(y=\frac{24}{x}\).
2. \(y=\frac{24}{16}=1.5\). Answer: \(y=1.5\).

Problem 18: Road trip

1. \(\frac{240}{40}=6\), \(\frac{240}{48}=5\), \(\frac{240}{60}=4\), \(\frac{240}{80}=3\). Times: 6, 5, 4 and 3 hours.
2. \(2.5=\frac{240}{r}\) gives \(r=\frac{240}{2.5}=96\). Speed: 96 mph.
3. The asymptote is \(t=0\). Driving faster always shortens the trip, but the time never reaches zero.

Problem 19: Cookie cost

1. \(C(20)=\frac{40}{20}=2\), \(C(60)=\frac{60}{60}=1\), \(C(120)=\frac{90}{120}=0.75\). Costs: $2.00, $1.00 and $0.75.
2. Split the fraction: \(\frac{0.5n}{n}+\frac{30}{n}=0.5+\frac{30}{n}\). The asymptote is \(C=0.5\): the average never drops below $0.50.
3. \(0.5n+30=0.6n\) gives \(30=0.1n\) and \(n=300\). Check: \(\frac{180}{300}=0.6\). Maya needs 300 cookies.

Problem 20: Bus rental

1. \(\frac{180}{12}=15\) and \(\frac{180}{20}=9\). Costs: $15 and $9 per student.
2. \(6=\frac{180}{n}\) gives \(n=30\). 30 students are needed.

Algebra 2 Rational Worksheet Answer Key: problems 21–22

Problem 21: Bike pump

1. \(k=PV=15\cdot 90=1350\), so \(P=\frac{1350}{V}\).
2. \(P=\frac{1350}{54}=25\). Pressure: 25 psi.
3. \(45=\frac{1350}{V}\) gives \(V=30\). Volume: 30 cubic inches.

Problem 22: Juice mixture

1. \(c(0)=\frac{2}{10}=20\%\), \(c(10)=\frac{12}{20}=60\%\), \(c(30)=\frac{32}{40}=80\%\).
2. \(\frac{2+x}{10+x}=\frac{1}{2}\) gives \(4+2x=10+x\), so \(x=6\). Check: \(\frac{8}{16}=\frac{1}{2}\). Ethan adds 6 gallons.
3. Equal degrees with leading coefficients 1 and 1 give \(c=1\), or 100%. The mixture always contains the original 8 gallons of non-concentrate liquid, so it is never pure concentrate.

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