Precalculus Exam 2 Study Guidance

Q1. Let $f(x) = -x^2 - x$. Find $f(-a)$.

Background

Topic: Function Evaluation and Substitution

This question tests your ability to substitute a value (in this case, $-a$) into a function and simplify the result.

Key Terms and Formulas

• Function evaluation: To find $f(k)$, replace every $x$ in $f(x)$ with $k$.

• Squaring a negative: $(-a)^2 = a^2$

Step-by-Step Guidance

1. Start by writing the function: $f(x) = -x^2 - x$.

2. Substitute $x = -a$ into the function: $f(-a) = -(-a)^2 - (-a)$.

3. Recall that $(-a)^2 = a^2$, and $-(-a) = a$.

4. Simplify the expression by applying these rules, but do not combine the terms yet.

Try solving on your own before revealing the answer!

Final Answer: D. $-a^2 - a$

Substituting $-a$ into the function gives $-(-a)^2 - (-a) = -a^2 + a$. But since the function is $-x^2 - x$, the correct simplification is $-a^2 - a$.

Q2. Find the $x$-intercepts for $g(x) = x^3 - x$.

Background

Topic: Finding $x$-intercepts (Roots/Zeros) of a Function

This question tests your ability to find the values of $x$ where the function equals zero.

Key Terms and Formulas

• $x$-intercept: A point where $y = 0$, so set $g(x) = 0$ and solve for $x$.

• Factoring: $x^3 - x = x(x^2 - 1) = x(x-1)(x+1)$

Step-by-Step Guidance

1. Set $g(x) = 0$: $x^3 - x = 0$.

2. Factor the expression: $x^3 - x = x(x^2 - 1)$.

3. Factor further: $x^2 - 1 = (x-1)(x+1)$, so $x(x-1)(x+1) = 0$.

4. Set each factor equal to zero to find the $x$-values where the function crosses the $x$-axis.

Try solving on your own before revealing the answer!

Final Answer: B. III (0, 0), (1, 0), and (−1, 0)

The solutions are $x = 0$, $x = 1$, and $x = -1$, so the $x$-intercepts are at these points.

Q3. Which of the following are functions?

Background

Topic: Definition of a Function (Vertical Line Test, Explicit/Implicit Relations)

This question tests your understanding of what makes a relation a function: each input ($x$) must correspond to exactly one output ($y$).

Key Terms and Formulas

• Function: For every $x$, there is only one $y$.

• Vertical Line Test: If a vertical line crosses the graph more than once, it is not a function.

Step-by-Step Guidance

1. For each equation, try to solve for $y$ in terms of $x$.

2. Check if, for every $x$, there is only one possible $y$ value.

3. For equations involving $y^2$, consider whether both positive and negative $y$ values are possible for a given $x$.

4. Identify which options meet the definition of a function.

Try solving on your own before revealing the answer!

Final Answer: B. I and III only

Only I and III define $y$ as a function of $x$; the others can give multiple $y$ values for a single $x$.

Q4. Suppose we perform the following transformations to the graph of $f$ in the indicated order: 1) reflection over the $y$-axis, 2) vertical stretch by a factor of 3, 3) shift down 2 units. What is the equation of the transformed graph?

Background

Topic: Transformations of Functions

This question tests your understanding of how to apply multiple transformations to a function and write the resulting equation.

Key Terms and Formulas

• Reflection over $y$-axis: Replace $x$ with $-x$.

• Vertical stretch by $a$: Multiply the function by $a$.

• Shift down $k$ units: Subtract $k$ from the function.

Step-by-Step Guidance

1. Start with $f(x)$.

2. Apply the reflection over the $y$-axis: $f(-x)$.

3. Apply the vertical stretch: $3f(-x)$.

4. Apply the downward shift: $3f(-x) - 2$.

Try solving on your own before revealing the answer!

Final Answer: A. $g(x) = 3f(-x) - 2$

Each transformation is applied in the correct order to get the final equation.

Q5. Suppose we perform the following transformations to the graph of $f(x) = |x|$ in the indicated order: 1) shift left 2 units, 2) reflection over the $x$-axis, 3) vertical stretch by a factor of 3, 4) shift up 1 unit. What is the equation of the transformed graph?

Background

Topic: Transformations of Absolute Value Functions

This question tests your ability to apply multiple transformations to the absolute value function and write the resulting equation.

Key Terms and Formulas

• Shift left $h$ units: Replace $x$ with $x + h$.

• Reflection over $x$-axis: Multiply the function by $-1$.

• Vertical stretch by $a$: Multiply the function by $a$.

• Shift up $k$ units: Add $k$ to the function.

Step-by-Step Guidance

1. Start with $f(x) = |x|$.

2. Shift left 2 units: $|x + 2|$.

3. Reflect over the $x$-axis: $-|x + 2|$.

4. Vertical stretch by 3: $-3|x + 2|$.

5. Shift up 1 unit: $-3|x + 2| + 1$.

Try solving on your own before revealing the answer!

Final Answer: D. $g(x) = -3|x + 2| + 1$

Each transformation is applied in the correct order to get the final equation.

Q6. Which of the following points on a graph are NOT moved by a horizontal reflection? Select all that apply.

Background

Topic: Symmetry and Transformations

This question tests your understanding of how points are affected by a reflection over the $y$-axis (horizontal reflection).

Key Terms and Formulas

• Horizontal reflection: Reflects a point $(x, y)$ to $(-x, y)$.

• Invariant points: Points where $x = 0$ are unchanged by this reflection.

Step-by-Step Guidance

1. Consider each point $(x, y)$ and apply the transformation: $(x, y) \to (-x, y)$.

2. Identify which points have $x = 0$; these will not move under the reflection.

3. Check the $x$-coordinate of each option to determine if it remains unchanged.

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Final Answer: B. (0, 1) and D. (0, 0)

Points with $x = 0$ are not moved by a horizontal reflection.

Q7. What is the domain of $f(x) = \frac{x}{\sqrt{1 - x}}$?

Background

Topic: Domain of Functions Involving Square Roots and Fractions

This question tests your ability to find the domain of a function that includes a square root in the denominator.

Key Terms and Formulas

• Domain: The set of all real $x$ values for which the function is defined.

• Square root in denominator: The expression under the square root must be positive (not zero or negative), and the denominator cannot be zero.

Step-by-Step Guidance

1. Set the denominator $\sqrt{1 - x}$ and require $1 - x > 0$ (since the denominator cannot be zero or negative under the square root).

2. Solve the inequality $1 - x > 0$ for $x$.

3. Express the solution as an interval.

Try solving on your own before revealing the answer!

Final Answer: C. $(-\infty, 1)$

The function is defined for all $x$ less than 1.

Q8. Let $f(x) = x^2$ and $g(x) = \sqrt{x - 1}$. What is the domain of $f \circ g$?

Background

Topic: Domain of Composite Functions

This question tests your ability to find the domain of a composition of functions, considering the domains of both the inner and outer functions.

Key Terms and Formulas

• Composite function: $(f \circ g)(x) = f(g(x))$.

• Domain of $g(x)$: $x - 1 \geq 0$.

• Domain of $f(x)$: All real numbers (since $x^2$ is defined for all $x$).

Step-by-Step Guidance

1. Find the domain of $g(x) = \sqrt{x - 1}$ by setting $x - 1 \geq 0$.

2. Solve for $x$ to find the domain of $g(x)$.

3. Since $f(x) = x^2$ is defined for all real numbers, the domain of $f \circ g$ is the same as the domain of $g(x)$.

Try solving on your own before revealing the answer!

Final Answer: D. $[1, \infty)$

The composite function is defined for $x \geq 1$.

Q9. Let $f(x) = \frac{x}{x-1}$ and $g(x) = \frac{2}{x}$. Find $(f \circ g)(x)$ and simplify.

Background

Topic: Composition of Rational Functions

This question tests your ability to compose two rational functions and simplify the result.

Key Terms and Formulas

• Composite function: $(f \circ g)(x) = f(g(x))$.

• Substitution: Replace $x$ in $f(x)$ with $g(x)$.

Step-by-Step Guidance

1. Write $f(x) = \frac{x}{x-1}$ and $g(x) = \frac{2}{x}$.

2. Substitute $g(x)$ into $f(x)$: $f(g(x)) = \frac{g(x)}{g(x) - 1}$.

3. Plug in $g(x)$: $\frac{\frac{2}{x}}{\frac{2}{x} - 1}$.

4. Simplify the denominator: $\frac{2}{x} - 1 = \frac{2 - x}{x}$.

5. Write the entire expression as a single fraction and simplify, but do not perform the final simplification yet.

Try solving on your own before revealing the answer!

Final Answer: B. $\frac{2}{2 - x}$

After simplifying, the composition yields $\frac{2}{2 - x}$.

Q10. Find the inverse of $f(x) = \frac{2}{x + 3}$.

Background

Topic: Inverse Functions

This question tests your ability to find the inverse of a rational function by solving for $x$ in terms of $y$ and then switching variables.

Key Terms and Formulas

• Inverse function: If $y = f(x)$, solve for $x$ in terms of $y$, then replace $y$ with $x$ to get $f^{-1}(x)$.

Step-by-Step Guidance

1. Start with $y = \frac{2}{x + 3}$.

2. Multiply both sides by $(x + 3)$ to clear the denominator.

3. Isolate $x$ by rearranging the equation.

4. Once $x$ is in terms of $y$, switch $x$ and $y$ to write the inverse function.

Try solving on your own before revealing the answer!

Final Answer: C. $f^{-1}(x) = \frac{2}{x} - 3$

Solving for $x$ and switching variables gives the correct inverse function.