AP Precalculus Unit 1 practice questions

Question 1

A polynomial function $$p(x)$$ has standard form $$p(x) = -2x^4 + 5x^3 + 8x - 3$$. Which statement correctly describes the end behavior and provides valid mathematical reasoning?

Accepted Answer

As $$x 	o \infty$$, $$p(x) 	o -\infty$$, and as $$x 	o -\infty$$, $$p(x) 	o -\infty$$, because the leading term is $$-2x^4$$ with even degree and negative coefficient.

Answer

As $$x 	o \infty$$, $$p(x) 	o \infty$$, and as $$x 	o -\infty$$, $$p(x) 	o -\infty$$, because the degree is even and the leading coefficient is negative.

Answer

As $$x 	o \infty$$, $$p(x) 	o -\infty$$, and as $$x 	o -\infty$$, $$p(x) 	o \infty$$, because the leading coefficient is negative.

Answer

As $$x 	o \infty$$, $$p(x) 	o \infty$$, and as $$x 	o -\infty$$, $$p(x) 	o \infty$$, because the polynomial has four terms.

Question 2

A rational function $$f(x) = \frac{2x^2 + 3x - 2}{x^2 + 1}$$ has numerator and denominator degrees that are equal. Which statement correctly identifies the horizontal asymptote and provides sound mathematical reasoning?

Accepted Answer

The horizontal asymptote is $$y = 2$$ because the degrees of numerator and denominator are equal, so the asymptote is the ratio of leading coefficients: $$\frac{2}{1} = 2$$.

Answer

The horizontal asymptote is $$y = 0$$ because the numerator degree equals the denominator degree, which means the function approaches zero at infinity.

Answer

There is no horizontal asymptote because the numerator and denominator both have degree 2, which means the function grows without bound.

Answer

The horizontal asymptote is $$y = 3$$ because the numerator has a constant term of $$-2$$ and the denominator has a constant term of $$1$$, so the ratio is $$\frac{3}{1} = 3$$.

Question 3

The polynomial $$p(x) = (x-1)^2(x+3)$$ is given in factored form. Which statement correctly identifies the x-intercepts and explains what the repeated factor reveals?

Accepted Answer

The x-intercepts are $$x = 1$$ and $$x = -3$$. The factor $$(x-1)^2$$ indicates that the graph touches the x-axis at $$x = 1$$ without crossing, because the zero has even multiplicity.

Answer

The x-intercepts are $$x = 1$$ and $$x = -3$$. The factor $$(x-1)^2$$ indicates that the graph crosses the x-axis at $$x = 1$$ with a steep slope because the exponent is 2.

Answer

The x-intercepts are $$x = 1, 1, -3$$ because $$(x-1)$$ appears twice in the factorization, creating three distinct intercepts.

Answer

The x-intercepts are $$x = 1$$ and $$x = -3$$. The factor $$(x-1)^2$$ indicates a vertical asymptote at $$x = 1$$ because the factor is squared.

Question 4

A rational function $$g(x) = \frac{x^3 - 8}{x - 2}$$ can be rewritten by factoring the numerator as $$g(x) = \frac{(x-2)(x^2+2x+4)}{x-2}$$. Which statement correctly describes the simplified form and the function's behavior?

Accepted Answer

For $$x 
eq 2$$, the function simplifies to $$g(x) = x^2 + 2x + 4$$, and there is a hole at $$(2, 12)$$ because the factor $$(x-2)$$ cancels and $$2^2 + 2(2) + 4 = 12$$.

Answer

For $$x 
eq 2$$, the function simplifies to $$g(x) = x^2 + 2x + 4$$, and there is a vertical asymptote at $$x = 2$$ because the original denominator is zero there.

Answer

The function cannot be simplified because the numerator and denominator have different degrees, so there is a vertical asymptote at $$x = 2$$.

Answer

For $$x 
eq 2$$, the function simplifies to $$g(x) = x^2 + 2x + 4$$, and the hole is at $$(2, 8)$$ because $$x - 2 = 0$$ when $$x = 2$$.

Question 5

Given $$f(x) = x^3 + 5x^2 + 8x + 4$$ and $$g(x) = x + 1$$, which of the following represents the quotient $$\frac{f(x)}{g(x)}$$ using polynomial long division?

Accepted Answer

$$x^2 + 4x + 4$$, with no remainder

Answer

$$x^2 + 4x + 4 + \frac{1}{x+1}$$

Answer

$$x^2 + 5x + 8$$, with remainder 4

Answer

$$x^2 + 6x + 2$$, with no remainder

Question 6

A rational function is defined as $$h(x) = \frac{2x^3 - 3x^2 - 8x + 12}{x - 2}$$. Using polynomial long division, which expression represents $$h(x)$$ and identifies any slant asymptote?

Accepted Answer

$$h(x) = 2x^2 + x - 6$$, with no slant asymptote

Answer

$$h(x) = 2x^2 + x - 6 + \frac{0}{x-2}$$, with slant asymptote $$y = 2x^2 + x - 6$$

Answer

$$h(x) = 2x^2 - x + 6$$, with slant asymptote $$y = 2x^2 - x$$

Answer

$$h(x) = 2x + 1 + \frac{10}{x-2}$$, with slant asymptote $$y = 2x + 1$$

Question 7

Given $$p(x) = x^4 - 1$$ and $$q(x) = x^2 - 1$$, which expression represents the quotient $$\frac{p(x)}{q(x)}$$ after polynomial long division and simplification?

Accepted Answer

$$x^2 + 1$$, with domain $$x 
eq \pm 1$$

Answer

$$x^2 + 1$$, with no domain restrictions

Answer

$$x^2 - 1$$, with domain $$x 
eq \pm 1$$

Answer

$$\frac{x^4 - 1}{x^2 - 1}$$ cannot be simplified by polynomial long division

Question 8

A rational function $$r(x) = \frac{x^3 + 2x^2 - 5x - 6}{x + 3}$$ is divided using polynomial long division. Which of the following correctly represents the result and identifies the type of discontinuity at $$x = -3$$?

Accepted Answer

$$r(x) = x^2 - x - 2$$, with a removable discontinuity (hole) at $$x = -3$$

Answer

$$r(x) = x^2 - x - 2 + \frac{0}{x+3}$$, with a vertical asymptote at $$x = -3$$

Answer

$$r(x) = x^2 + x - 2$$, with a removable discontinuity at $$x = -3$$

Answer

$$r(x) = x^2 - x - 2 + \frac{5}{x+3}$$, with a vertical asymptote at $$x = -3$$

Question 9

Given $$f(x) = x^4 + 3x^3 - x - 3$$ and $$g(x) = x + 3$$, polynomial long division produces quotient $$q(x)$$ and remainder $$r$$. Which expression correctly represents $$f(x)$$ in the form $$f(x) = g(x) \cdot q(x) + r$$?

Accepted Answer

$$f(x) = (x+3)(x^3 - 1) + 0$$

Answer

$$f(x) = (x+3)(x^3 + 1) + 0$$

Answer

$$f(x) = (x+3)(x^3 - 1) - 6$$

Answer

$$f(x) = (x+3)(x^2 + 1) + (x-3)$$

Question 10

The rational function $$h(x) = \frac{x^2 - 2x - 3}{x^2 - 9}$$ can be simplified by canceling a common factor. Which of the following represents the simplified form and identifies the hole?

Accepted Answer

Simplified: $$h(x) = \frac{x + 1}{x + 3}$$; hole at $$(3, \frac{2}{3})$$

Answer

Simplified: $$h(x) = \frac{x - 1}{x - 3}$$; hole at $$(3, 1)$$

Answer

Simplified: $$h(x) = \frac{x + 1}{x + 3}$$; hole at $$(-3, 0)$$

Answer

Simplified: $$h(x) = \frac{x + 1}{x + 3}$$; no hole exists

AP Precalculus Unit 1 Practice Test

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