12 multiple-choice questions, progressively harder.
What is the horizontal asymptote of f(x)=4−3x26x2+xf(x) = \dfrac{4 - 3x^2}{6x^2 + x}f(x)=6x2+x4−3x2?
Solution
Correct answer: A
The degrees match at 222, so take the ratio of leading coefficients: −3-3−3 on top, 666 on the bottom.
y=−36=−12y = \frac{-3}{6} = -\frac{1}{2}y=6−3=−21
Find the coordinates of the hole in f(x)=x2−x−12x−4f(x) = \dfrac{x^2 - x - 12}{x - 4}f(x)=x−4x2−x−12.
Correct answer: B
Factor and cancel.
x2−x−12x−4=(x−4)(x+3)x−4=x+3(x≠4)\frac{x^2 - x - 12}{x - 4} = \frac{(x - 4)(x + 3)}{x - 4} = x + 3 \quad (x \neq 4)x−4x2−x−12=x−4(x−4)(x+3)=x+3(x=4)
The hole is at x=4x = 4x=4 with height 4+3=74 + 3 = 74+3=7, so it is (4,7)(4, 7)(4,7).
Solve x+3x−1=2\dfrac{x + 3}{x - 1} = 2x−1x+3=2.
Correct answer: C
Multiply both sides by x−1x - 1x−1.
x+3=2(x−1) ⇒ x+3=2x−2 ⇒ x=5x + 3 = 2(x - 1) \;\Rightarrow\; x + 3 = 2x - 2 \;\Rightarrow\; x = 5x+3=2(x−1)⇒x+3=2x−2⇒x=5
Checking, 5+35−1=84=2\frac{5 + 3}{5 - 1} = \frac{8}{4} = 25−15+3=48=2, so x=5x = 5x=5 is valid.
Solve 5x−2=1x+2\dfrac{5}{x - 2} = \dfrac{1}{x + 2}x−25=x+21.
Cross-multiply.
5(x+2)=1(x−2) ⇒ 5x+10=x−2 ⇒ 4x=−12 ⇒ x=−35(x + 2) = 1(x - 2) \;\Rightarrow\; 5x + 10 = x - 2 \;\Rightarrow\; 4x = -12 \;\Rightarrow\; x = -35(x+2)=1(x−2)⇒5x+10=x−2⇒4x=−12⇒x=−3
Checking, 5−5=−1\frac{5}{-5} = -1−55=−1 and 1−1=−1\frac{1}{-1} = -1−11=−1, so x=−3x = -3x=−3 is valid.
Solve x2x−1=1x−1\dfrac{x^2}{x - 1} = \dfrac{1}{x - 1}x−1x2=x−11.
Correct answer: D
x2=1 ⇒ x=1 or x=−1x^2 = 1 \;\Rightarrow\; x = 1 \text{ or } x = -1x2=1⇒x=1 or x=−1
The candidate x=1x = 1x=1 makes the denominator zero, so it is extraneous. Only x=−1x = -1x=−1 survives.
How many vertical asymptotes does f(x)=x+1x3−4xf(x) = \dfrac{x + 1}{x^3 - 4x}f(x)=x3−4xx+1 have?
Factor the denominator; the numerator cancels nothing.
x3−4x=x(x−2)(x+2)=0 ⇒ x=0, 2, −2x^3 - 4x = x(x - 2)(x + 2) = 0 \;\Rightarrow\; x = 0, \ 2, \ -2x3−4x=x(x−2)(x+2)=0⇒x=0, 2, −2
All three factors survive, so there are three vertical asymptotes.
Find the hole in the graph of f(x)=x2−1x−1f(x) = \dfrac{x^2 - 1}{x - 1}f(x)=x−1x2−1.
x2−1x−1=(x−1)(x+1)x−1=x+1(x≠1)\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \quad (x \neq 1)x−1x2−1=x−1(x−1)(x+1)=x+1(x=1)
The hole is at x=1x = 1x=1 with height 1+1=21 + 1 = 21+1=2, so it is (1,2)(1, 2)(1,2).
At what point do the asymptotes of f(x)=−3x+4+2f(x) = \dfrac{-3}{x + 4} + 2f(x)=x+4−3+2 cross?
Match ax−h+k\tfrac{a}{x - h} + kx−ha+k: writing x+4x + 4x+4 as x−(−4)x - (-4)x−(−4) gives h=−4h = -4h=−4 and k=2k = 2k=2.
(h,k)=(−4,2)(h, k) = (-4, 2)(h,k)=(−4,2)
The asymptotes x=−4x = -4x=−4 and y=2y = 2y=2 cross at (−4,2)(-4, 2)(−4,2).
Which rule produces the reciprocal graph flipped so its branches lie in the 2nd and 4th quadrants?
A negative coefficient reflects the hyperbola across its horizontal asymptote.
y=−1xy = -\frac{1}{x}y=−x1
Now positive inputs give negative outputs and vice versa, moving the branches to the 2nd and 4th quadrants.
What is the horizontal asymptote of f(x)=6x3−x3x3+2f(x) = \dfrac{6x^3 - x}{3x^3 + 2}f(x)=3x3+26x3−x?
The degrees match at 333, so use the ratio of leading coefficients.
y=63=2y = \frac{6}{3} = 2y=36=2
How many holes does the graph of f(x)=(x−1)(x+2)(x−1)(x+2)(x−5)f(x) = \dfrac{(x - 1)(x + 2)}{(x - 1)(x + 2)(x - 5)}f(x)=(x−1)(x+2)(x−5)(x−1)(x+2) have?
Cancel every shared factor.
(x−1)(x+2)(x−1)(x+2)(x−5)=1x−5(x≠1, x≠−2)\frac{(x - 1)(x + 2)}{(x - 1)(x + 2)(x - 5)} = \frac{1}{x - 5} \quad (x \neq 1, \ x \neq -2)(x−1)(x+2)(x−5)(x−1)(x+2)=x−51(x=1, x=−2)
Both x−1x - 1x−1 and x+2x + 2x+2 cancel, giving two holes; x−5x - 5x−5 survives as an asymptote.
What is the vertical asymptote of f(x)=x−5x2−25f(x) = \dfrac{x - 5}{x^2 - 25}f(x)=x2−25x−5?
x−5x2−25=x−5(x−5)(x+5)=1x+5(x≠5)\frac{x - 5}{x^2 - 25} = \frac{x - 5}{(x - 5)(x + 5)} = \frac{1}{x + 5} \quad (x \neq 5)x2−25x−5=(x−5)(x+5)x−5=x+51(x=5)
The factor x−5x - 5x−5 cancels (a hole at x=5x = 5x=5), so the only vertical asymptote is x=−5x = -5x=−5.
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