Rational Functions Worksheet Answer Key – Algebra 2

Problem 1 Finding the domain

a) x − 3 = 0 at x = 3: all real numbers except 3.

b) x² − 16 = (x − 4)(x + 4): all real numbers except 4 and −4.

c) 2x + 5 = 0 at x = –52: all real numbers except –52.

Problem 2 Asymptotes from the equation

a) The form a/(x − h) + k has h = −4 and k = −3: x = −4 and y = −3.

b) h = 5 and k = 6: x = 5 and y = 6.

Problem 3 Describing transformations

Replacing x by x + 3 shifts the graph left 3. Subtracting 5 shifts it down 5.

Left 3 and down 5; the asymptotes move to x = −3 and y = −5.

Problem 4 Reading a graph

The branches approach the dashed lines x = 1 and y = 2.

Asymptotes x = 1 and y = 2; domain: all reals except 1; range: all reals except 2

Problem 5 Inverse variation constant

k = xy = 3 · 12 = 36, so y = 36x.

When x = 9: y = 36/9 = 4.

y = 36x; y = 4

Problem 6 A hole in the graph

(x − 6)(x + 6)x − 6 = x + 6 for x ≠ 6. At x = 6 the simplified expression gives 12.

Hole at (6, 12)

Problem 7 A hole and an asymptote

The factor x + 2 cancels, leaving 1x − 5. The hole is at x = −2, with height 1/(−2 − 5) = –17.

Hole at (−2, –17); vertical asymptote x = 5

Problem 8 Horizontal asymptotes by degree

a) Equal degrees: ratio of leading coefficients 3/1: y = 3.

b) Numerator degree 1 < denominator degree 2: y = 0.

c) Numerator degree 3 > denominator degree 2: none.

Problem 9 Writing an equation from features

h = −3 and k = 2, so y = ax + 3 + 2. At (0, 3): 3 = a/3 + 2, so a = 3.

y = 3x + 3 + 2

Problem 10 Intercepts and asymptotes

x-intercept: 2x − 6 = 0 gives x = 3. y-intercept: y = −6/1 = −6.

Vertical asymptote where x + 1 = 0. Equal degrees: horizontal asymptote y = 2/1 = 2.

(3, 0); (0, −6); x = −1; y = 2

Problem 11 Sketching a rational function

Asymptotes: x = −1 and y = −3. Since a = 2 > 0, the branches are above-right and below-left of the crossing point (−1, −3).

Points: x = 0 gives −1; x = 1 gives −2; x = −2 gives −5; x = −3 gives −4.

Asymptotes x = −1 and y = −3

Problem 12 Is it inverse variation?

Products: 4 · 15 = 60, 5 · 12 = 60, 6 · 10 = 60, 10 · 6 = 60. The product is constant, so y varies inversely with x.

y = 60x. At x = 15: y = 4.

y = 60x; y = 4

Problem 13 Rewriting in standard form

3x + 5 = 3(x + 2) − 1, so 3x + 5x + 2 = 3 − 1x + 2.

So y = –1x + 2 + 3.

x = −2 and y = 3

Problem 14 Domain and hole from a graph

Factor: (x − 3)(x + 2)x − 3 = x + 2 for x ≠ 3. The hole is at x = 3, with height 3 + 2 = 5.

Hole (3, 5); domain all reals except 3; line y = x + 2

Problem 15 Solving for an input

f(x) = 5: 6/(x − 1) = 3, so x − 1 = 2 and x = 3.

f(x) = 0: 6/(x − 1) = −2, so x − 1 = −3 and x = −2.

x = 3; x = −2

Problem 16 A road trip

At 60 mph: t = 240/60 = 4 hours.

For t = 3.2: r = 240/3.2 = 75 mph.

As r grows, t gets closer to 0 but never reaches it, so the horizontal asymptote is t = 0.

4 hours; 75 mph; the time approaches 0

Problem 17 Diluting a solution

210 + x = 0.1, so 10 + x = 20 and x = 10.

Domain: x ≥ 0. Horizontal asymptote: c = 0, since the numerator has the smaller degree.

10 liters of water; asymptote c = 0: the acid fraction approaches 0 but never reaches it

Problem 18 Average cost of a food truck

3n + 1,200n = 3 + 1,200n.

Set 3 + 1,200/n = 5: 1,200/n = 2, so n = 600.

As n grows, 1,200/n approaches 0, so A approaches 3.

A(n) = 3 + 1,200n; n = 600; horizontal asymptote A = 3

Problem 19 All the features of one function

Factor: (x − 4)(x + 3)(x − 4)(x + 2). The factor x − 4 cancels, leaving x + 3x + 2.

Hole at x = 4: height 7/6. Vertical asymptote: x = −2. Horizontal asymptote: equal degrees, y = 1.

x-intercept: x + 3 = 0 gives x = −3. y-intercept: f(0) = −12/−8 = 32.

Hole (4, 76); x = −2; y = 1; (−3, 0); (0, 32)

Problem 20 Designing a rational function

The denominator needs the factors (x − 4) for the asymptote and (x + 3) for the hole, and the numerator must also contain (x + 3). For the x-intercept 7 include (x − 7).

Equal degrees with a ratio of leading coefficients 5 requires a factor 5 in the numerator.

f(x) = 5(x − 7)(x + 3)(x − 4)(x + 3). The simplified form is 5(x − 7)x − 4; at x = −3 it equals 5(−10)/(−7) = 507.

f(x) = 5(x − 7)(x + 3)(x − 4)(x + 3); hole at (−3, 507)