12 multiple-choice questions, progressively harder.
What is the domain of f(x)=2x−1x2−5x+6f(x) = \dfrac{2x - 1}{x^2 - 5x + 6}f(x)=x2−5x+62x−1?
Solution
Correct answer: C
Factor the denominator and set each factor to zero.
x2−5x+6=(x−2)(x−3)=0 ⇒ x=2 or x=3x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \;\Rightarrow\; x = 2 \text{ or } x = 3x2−5x+6=(x−2)(x−3)=0⇒x=2 or x=3
The domain is every real number except 222 and 333.
What is the horizontal asymptote of f(x)=1−4x22x2+xf(x) = \dfrac{1 - 4x^2}{2x^2 + x}f(x)=2x2+x1−4x2?
Correct answer: D
The degrees match at 222, so take the ratio of leading coefficients: −4-4−4 on top, 222 on the bottom.
y=−42=−2y = \frac{-4}{2} = -2y=2−4=−2
How many x-intercepts does f(x)=x2−x−6x+4f(x) = \dfrac{x^2 - x - 6}{x + 4}f(x)=x+4x2−x−6 have?
Set the numerator to zero.
x2−x−6=(x−3)(x+2)=0 ⇒ x=3 or x=−2x^2 - x - 6 = (x - 3)(x + 2) = 0 \;\Rightarrow\; x = 3 \text{ or } x = -2x2−x−6=(x−3)(x+2)=0⇒x=3 or x=−2
Neither makes the denominator zero, so there are two x-intercepts.
Solve xx−3=6x−3−1\dfrac{x}{x - 3} = \dfrac{6}{x - 3} - 1x−3x=x−36−1.
Correct answer: B
Multiply every term by x−3x - 3x−3.
x=6−(x−3) ⇒ x=9−x ⇒ 2x=9 ⇒ x=92x = 6 - (x - 3) \;\Rightarrow\; x = 9 - x \;\Rightarrow\; 2x = 9 \;\Rightarrow\; x = \tfrac{9}{2}x=6−(x−3)⇒x=9−x⇒2x=9⇒x=29
Since x=92≠3x = \tfrac{9}{2} \neq 3x=29=3, the denominator is nonzero and the solution is valid.
Which statement is true about the graph of f(x)=x2+2xx2−4f(x) = \dfrac{x^2 + 2x}{x^2 - 4}f(x)=x2−4x2+2x?
Factor and cancel.
x2+2xx2−4=x(x+2)(x−2)(x+2)=xx−2(x≠−2)\frac{x^2 + 2x}{x^2 - 4} = \frac{x(x + 2)}{(x - 2)(x + 2)} = \frac{x}{x - 2} \quad (x \neq -2)x2−4x2+2x=(x−2)(x+2)x(x+2)=x−2x(x=−2)
The factor x+2x + 2x+2 cancels (hole at x=−2x = -2x=−2), x−2x - 2x−2 survives (asymptote at x=2x = 2x=2), and equal degrees give y=1y = 1y=1.
Solve 1x−1=4\dfrac{1}{x - 1} = 4x−11=4.
Correct answer: A
Multiply both sides by x−1x - 1x−1.
1=4(x−1) ⇒ 1=4x−4 ⇒ 4x=5 ⇒ x=541 = 4(x - 1) \;\Rightarrow\; 1 = 4x - 4 \;\Rightarrow\; 4x = 5 \;\Rightarrow\; x = \tfrac{5}{4}1=4(x−1)⇒1=4x−4⇒4x=5⇒x=45
Since x=54≠1x = \tfrac{5}{4} \neq 1x=45=1, the solution is valid.
What is the domain of f(x)=x−7x(x−1)(x+3)f(x) = \dfrac{x - 7}{x(x - 1)(x + 3)}f(x)=x(x−1)(x+3)x−7?
The denominator is zero where any factor is zero.
x=0,x−1=0,x+3=0 ⇒ x=0, 1, −3x = 0, \quad x - 1 = 0, \quad x + 3 = 0 \;\Rightarrow\; x = 0, \ 1, \ -3x=0,x−1=0,x+3=0⇒x=0, 1, −3
The domain excludes all three.
Find the coordinates of the hole in f(x)=2x2−8x−2f(x) = \dfrac{2x^2 - 8}{x - 2}f(x)=x−22x2−8.
Factor out the 222 and cancel.
2x2−8x−2=2(x−2)(x+2)x−2=2(x+2)(x≠2)\frac{2x^2 - 8}{x - 2} = \frac{2(x - 2)(x + 2)}{x - 2} = 2(x + 2) \quad (x \neq 2)x−22x2−8=x−22(x−2)(x+2)=2(x+2)(x=2)
The hole is at x=2x = 2x=2 with height 2(2+2)=82(2 + 2) = 82(2+2)=8, so it is (2,8)(2, 8)(2,8).
How does f(x)=x4x2+1f(x) = \dfrac{x^4}{x^2 + 1}f(x)=x2+1x4 behave for very large xxx?
The numerator degree 444 exceeds the denominator degree 222, so n>mn > mn>m.
x4x2=x2\frac{x^4}{x^2} = x^2x2x4=x2
So for very large xxx the expression behaves like x2x^2x2, which grows without bound, and a top-heavy fraction has no horizontal asymptote.
Solve 2x+3x=1\dfrac{2}{x} + \dfrac{3}{x} = 1x2+x3=1.
Combine the like fractions, then multiply by xxx.
5x=1 ⇒ 5=x ⇒ x=5\frac{5}{x} = 1 \;\Rightarrow\; 5 = x \;\Rightarrow\; x = 5x5=1⇒5=x⇒x=5
Since x=5≠0x = 5 \neq 0x=5=0, the solution is valid.
The graph of y=1x−3+2y = \dfrac{1}{x - 3} + 2y=x−31+2 approaches which two lines?
In the form ax−h+k\tfrac{a}{x - h} + kx−ha+k, read h=3h = 3h=3 and k=2k = 2k=2.
x=3andy=2x = 3 \quad \text{and} \quad y = 2x=3andy=2
The vertical asymptote is x=3x = 3x=3 and the horizontal asymptote is y=2y = 2y=2.
Solve x+6x=5x + \dfrac{6}{x} = 5x+x6=5.
Multiply every term by xxx to clear the fraction.
x2+6=5x ⇒ x2−5x+6=0 ⇒ (x−2)(x−3)=0x^2 + 6 = 5x \;\Rightarrow\; x^2 - 5x + 6 = 0 \;\Rightarrow\; (x - 2)(x - 3) = 0x2+6=5x⇒x2−5x+6=0⇒(x−2)(x−3)=0
Both x=2x = 2x=2 and x=3x = 3x=3 keep x≠0x \neq 0x=0, so both are valid.
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