College Algebra Questions With Answers
Sample 1

Using the logarithmic identity \( b^{\log_b(x)} = x \), we have:

\[ 9^{\log_{9}(4)} = 4 \]

Correct Answer: B

The argument of a logarithm must be strictly positive (\( x > 0 \)). Therefore, \(\log_{3}(-5)\) is undefined in the real number system.

Correct Answer: D

Let's analyze the properties of the quadratic function \( f(x) = -2x^2 + 8x - 4 \):

• The x-coordinate of the vertex is \( x = -\frac{b}{2a} = -\frac{8}{2(-2)} = 2 \).
• The maximum value occurs at \( x = 2 \): \( f(2) = -2(2)^2 + 8(2) - 4 = -8 + 16 - 4 = 4 \) (so A is false).
• Since the leading coefficient is \(-2 < 0\), the graph opens downward (so B is false).
• The discriminant is \( \Delta = 8^2 - 4(-2)(-4) = 64 - 32 = 32 > 0 \), meaning there are two x-intercepts (so C is false).
• Since it is a parabola, it is not a one-to-one function on its natural domain (so D is true).

Correct Answer: D

We want to find \( x \) such that \( f(x) = -3 \):

\[ 5 - 2^x = -3 \implies 2^x = 8 \implies x = 3 \]

Thus, \( f^{-1}(-3) = 3 \).

Correct Answer: C

Rewrite the logarithmic equation in exponential form:

\[ x^{1/4} = 3 \implies (x^{1/4})^4 = 3^4 \implies x = 81 \]

Correct Answer: A

Substitute \( x + 1 \) for \( x \) in the function:

\[ f(x + 1) = -(x + 1)^2 + 1 = -(x^2 + 2x + 1) + 1 = -x^2 - 2x - 1 + 1 = -x^2 - 2x \]

Correct Answer: B

Evaluate step by step:

\[ f(3) = 3 - 4 = -1 \] \[ (f \circ f)(3) = f(f(3)) = f(-1) = -1 - 4 = -5 \]

Correct Answer: C

Rewrite the natural logarithm equation in exponential form (base \(e\)):

\[ 3x - 2 = e^1 \implies 3x = e + 2 \implies x = \frac{2 + e}{3} \]

Correct Answer: B

Let \( u = x^2 + 1 \). The equation becomes:

\[ u^2 + 2u - 3 = 0 \implies (u + 3)(u - 1) = 0 \]

This gives \( u = -3 \) or \( u = 1 \):

• If \( x^2 + 1 = -3 \implies x^2 = -4 \) (no real solutions).
• If \( x^2 + 1 = 1 \implies x^2 = 0 \implies x = 0 \) (1 real solution).

Thus, there is only 1 real solution.

Correct Answer: A

• Translating 2 units to the right replaces \( x \) with \( x - 2 \): \( y = ((x - 2) - 2)^2 - 3 = (x - 4)^2 - 3 \).
• Translating 5 units up adds \( 5 \) to the function: \( y = (x - 4)^2 - 3 + 5 = (x - 4)^2 + 2 \).

Correct Answer: D

Since \( e^x > 0 \) for all real numbers \( x \), it follows that \(-e^x < 0\), and consequently \(-e^x - 2 < -2\). Therefore, the range is \((-\infty, -2)\).

Correct Answer: A

Determine the domain restrictions:

• The expression under the square root must be non-negative: \( x - 1 \ge 0 \implies x \ge 1 \).
• The denominator cannot be zero: \( x^2 - 9 \neq 0 \implies x \neq 3 \) and \( x \neq -3 \).

Combining these conditions (\( x \ge 1 \) and \( x \neq 3 \)) gives the interval \([1, 3) \cup (3, +\infty)\).

Correct Answer: C

Analyzing the curves:

• At \( x = 0 \): \( 2^0 = 1 \) and \(-0^2 + 2 = 2\) (parabola is above).
• At \( x = 1 \): \( 2^1 = 2 \) and \(-1^2 + 2 = 1\) (exponential is above). By the Intermediate Value Theorem, an intersection exists between \( x = 0 \) and \( x = 1 \).
• At \( x = -1 \): \( 2^{-1} = 0.5 \) and \(-(-1)^2 + 2 = 1\) (parabola is above).
• At \( x = -2 \): \( 2^{-2} = 0.25 \) and \(-(-2)^2 + 2 = -2\) (exponential is above). By the Intermediate Value Theorem, another intersection exists between \( x = -2 \) and \( x = -1 \).

Thus, there are 2 intersection points.

Correct Answer: C

Set \( y = \ln(x + 1) - 2 \) and solve for \( x \):

\[ y + 2 = \ln(x + 1) \implies e^{y + 2} = x + 1 \implies x = e^{y + 2} - 1 \]

Replacing \( y \) with \( x \) gives \( f^{-1}(x) = e^{x + 2} - 1 \).

Correct Answer: D

Factor the quadratic expression inside the square root:

\[ \sqrt{x^2 - 4x + 4} = \sqrt{(x - 2)^2} = |x - 2| \]

Correct Answer: C

Note: A quadratic expression with a positive leading coefficient (\( a = 1 > 0 \)) opens upward and has a minimum, not a maximum. However, the x-coordinate of the vertex (where the extremum occurs) is given by:

\[ x = -\frac{b}{2a} = -\frac{6}{2(1)} = -3 \]

Correct Answer: B

Combine the logarithmic terms using logarithm properties:

\[ \ln(3) - \ln(2) + \ln\left(\frac{1}{x}\right) = \ln\left(\frac{3}{2} \cdot \frac{1}{x}\right) = \ln\left(\frac{3}{2x}\right) \]

Applying the exponential function:

\[ e^{\ln\left(\frac{3}{2x}\right)} = \frac{3}{2x} \]

Correct Answer: A

For a rational function of the form \( \frac{ax + b}{cx + d} \), the horizontal asymptote is at \( y = \frac{a}{c} = \frac{1}{1} = 1 \). The function never attains this value, so the range excludes \( 1 \), yielding \((-\infty, 1) \cup (1, +\infty)\).

Correct Answer: B

The property \( \ln(a^b) = b \ln(a) \) requires the argument \( a > 0 \). While \((x-1)^2 > 0\) for all \( x \neq 1 \), the right-hand side contains \( \ln(x - 1)\), which strictly requires \( x - 1 > 0 \implies x > 1 \). Therefore, the identity holds for \( x \) in \((1, +\infty)\).

Correct Answer: D

Let's evaluate each statement:

• A: Since \( a = 1 > 0 \), the parabola opens upward and has a minimum value, not a maximum value (False / NOT true).
• B: A quadratic function's graph is a parabola, not a line (True).
• C: Discriminant \( \Delta = 2^2 - 4(1)(4) = 4 - 16 = -12 < 0 \), so there are no real x-intercepts (True).
• D: The y-intercept is \( f(0) = 4 \) (True).

Correct Answer: A