Precalculus Rational Functions and Asymptotes Quiz: 10 Questions with Explanations

A \(x=4\)B \(x=-4\)C \(y=-4\)D \(y=5\)

The denominator is 0 at x = -4. Vertical asymptotes are lines x = a.

A \(y=0\)B \(y=3\)C \(y=2\)D \(y=6\)

Equal degrees: 6 divided by 3 gives 2.

A A holeB A vertical asymptoteC An x-interceptD A horizontal asymptote

x – 1 cancels, so the graph is y = x + 1 with a hole at (1, 2).

A \(y=3\)B \(y=2\)C \(x=3\)D \(y=0\)

The dashed horizontal line is y = 2. The line x = 3 is the vertical asymptote.

A \(y=1\)B \(y=x+1\)C \(y=x\)D There is none

x^2 + 1 over x = x + 1/x, so the line is y = x.

A No solutionB \(x=1\)C \(x=-1\)D \(x=2\)

Solving gives x = 1, which makes the denominator 0, so it is extraneous.

A \(\frac{x-2}{(x-2)(x+1)}\)B \(\frac{x+1}{(x-2)(x+1)}\)C \(\frac{1}{(x-2)(x+1)}\)D \(\frac{x+1}{x-2}\)

The cancelled factor x + 1 gives the hole at -1, and the remaining x – 2 gives the asymptote.

A \((0,-2)\)B \((6,0)\)C \((0,2)\)D \((-3,0)\)

f(0) = -6/3 = -2. The value 6 gives the x-intercept.

A \((4,2)\)B \((2,0)\)C \((0,2)\)D \((2,4)\)

The open circle is at x = 2 on the line y = x + 2, so y = 4.

A \(x\neq 2\)B \(x\neq -2\)C All real numbersD \(x\neq\pm2\)

Both x = 2 and x = -2 make the denominator 0. The cancelled factor still removes -2 from the domain.