College Algebra Exam 1 Review: Step-by-Step Guidance

Q1. Given $f(x) = x^2 - 3x + 2$, find (a) $f(-2)$ and (b) $f(p + h)$

Background

Topic: Function Evaluation and Algebraic Manipulation

This question tests your ability to evaluate a function at a specific value and to substitute an expression into a function.

Key Terms and Formulas:

• Function notation: $f(x)$ means "the value of the function f at x"

• To evaluate $f(a)$, substitute $a$ for $x$ in the formula.

Step-by-Step Guidance

1. For part (a), substitute $x = -2$ into $f(x)$: $f(-2) = (-2)^2 - 3(-2) + 2$

2. Calculate each term: $(-2)^2$, $-3(-2)$, and $+2$.

3. Add the results together to get $f(-2)$.

4. For part (b), substitute $x = p + h$ into $f(x)$: $f(p + h) = (p + h)^2 - 3(p + h) + 2$

5. Expand $(p + h)^2$ and $-3(p + h)$, then combine like terms.

Try solving on your own before revealing the answer!

Final Answer:

(a) $f(-2) = 12$

(b) $f(p + h) = p^2 + 2ph + h^2 - 3p - 3h + 2$

For (a), substitute and simplify. For (b), expand and combine like terms.

Q2. Given $f(x) = x^2 - 4$, find $f(x + h) - f(x)$

Background

Topic: Difference Quotient and Function Operations

This question tests your ability to compute the difference between a function evaluated at $x + h$ and at $x$.

Key Terms and Formulas:

• Difference: $f(x + h) - f(x)$

• Expand and simplify expressions.

Step-by-Step Guidance

1. Write $f(x + h) = (x + h)^2 - 4$

2. Write $f(x) = x^2 - 4$

3. Subtract: $f(x + h) - f(x) = [(x + h)^2 - 4] - [x^2 - 4]$

4. Expand $(x + h)^2$ to $x^2 + 2xh + h^2$

5. Combine like terms and simplify the expression.

Try solving on your own before revealing the answer!

Final Answer:

$f(x + h) - f(x) = 2xh + h^2$

The $x^2$ and $-4$ terms cancel, leaving the expanded terms.

Q3. Find the domain of the function:

• (a) $f(x) = \frac{1}{3 - x}$

• (b) $f(x) = \frac{2x - 3}{x^2 + 5x - 3}$

Background

Topic: Domain of Rational Functions

This question tests your ability to find the set of all real numbers for which the function is defined (i.e., where the denominator is not zero).

Key Terms and Formulas:

• Domain: All real values of $x$ for which the function is defined.

• For rational functions, exclude values where the denominator is zero.

Step-by-Step Guidance

1. For (a), set the denominator $3 - x \neq 0$ and solve for $x$.

2. For (b), set the denominator $x^2 + 5x - 3 \neq 0$ and solve for $x$.

3. Find the values of $x$ that make the denominator zero (for (b), use the quadratic formula).

4. Write the domain as all real numbers except those values.

Try solving on your own before revealing the answer!

Final Answer:

(a) Domain: $x \neq 3$

(b) Domain: $x \neq \frac{-5 \pm \sqrt{25 + 12}}{2}$, or $x \neq \frac{-5 \pm \sqrt{37}}{2}$

Exclude values that make the denominator zero.

Q4. Graph the following piecewise function:

• $f(x) = -x$ for $x < -4$

• $f(x) = x^2 + 1$ for $-4 \leq x < 2$

• $f(x) = 1$ for $x \geq 2$

Background

Topic: Piecewise Functions and Graphing

This question tests your ability to interpret and graph a function defined by different expressions over different intervals.

Key Terms and Formulas:

• Piecewise function: A function defined by different formulas for different parts of its domain.

• Graph each segment according to its interval.

Step-by-Step Guidance

1. Identify the intervals and corresponding expressions.

2. For $x < -4$, graph $f(x) = -x$ (a straight line).

3. For $-4 \leq x < 2$, graph $f(x) = x^2 + 1$ (a parabola segment).

4. For $x \geq 2$, graph $f(x) = 1$ (a horizontal line).

5. Mark endpoints and check for open/closed circles at interval boundaries.

Try solving on your own before revealing the answer!

Final Answer:

Graph consists of:

• Line $f(x) = -x$ for $x < -4$

• Parabola segment $f(x) = x^2 + 1$ for $-4 \leq x < 2$

• Horizontal line $f(x) = 1$ for $x \geq 2$

At $x = -4$, use a closed circle for $x^2 + 1$; at $x = 2$, use a closed circle for $f(x) = 1$.

Q5. Find the equation of the line that passes through $(-5, 4)$ and $(3, -2)$. Give the answer in slope y-intercept form.

Background

Topic: Linear Equations and Slope-Intercept Form

This question tests your ability to find the equation of a line given two points.

Key Terms and Formulas:

• Slope formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$

• Slope-intercept form: $y = mx + b$

Step-by-Step Guidance

1. Label the points: $(-5, 4)$ as $(x_1, y_1)$ and $(3, -2)$ as $(x_2, y_2)$.

2. Calculate the slope $m$ using the formula.

3. Use the slope and one point to solve for $b$ in $y = mx + b$.

4. Write the final equation in $y = mx + b$ form.

Try solving on your own before revealing the answer!

Final Answer:

Slope $m = \frac{-2 - 4}{3 - (-5)} = \frac{-6}{8} = -\frac{3}{4}$

Equation: $y = -\frac{3}{4}x + \frac{7}{4}$

Plug the slope and one point into the formula to solve for $b$.

Q6. Find the equation of the line passing through $(-1, 3)$ that is parallel to the line $2x + 6y = -6$. Also, find the equation of the line passing through $(-1, 3)$ that is perpendicular to $2x + 6y = -6$. Give both answers in slope y-intercept form.

Background

Topic: Parallel and Perpendicular Lines

This question tests your ability to find equations of lines parallel and perpendicular to a given line, passing through a specific point.

Key Terms and Formulas:

• Parallel lines have the same slope.

• Perpendicular lines have slopes that are negative reciprocals.

• Convert $2x + 6y = -6$ to slope-intercept form to find the slope.

• Slope-intercept form: $y = mx + b$

Step-by-Step Guidance

1. Rewrite $2x + 6y = -6$ in $y = mx + b$ form to find the slope.

2. For the parallel line, use the same slope and the point $(-1, 3)$ to solve for $b$.

3. For the perpendicular line, use the negative reciprocal of the slope and the point $(-1, 3)$ to solve for $b$.

4. Write both equations in $y = mx + b$ form.

Try solving on your own before revealing the answer!

Final Answer:

Original line slope: $m = -\frac{1}{3}$

Parallel line: $y = -\frac{1}{3}x + \frac{8}{3}$

Perpendicular line: $y = 3x + 6$

Use the point-slope formula and solve for $b$ in each case.

Q7. Find the equation of the circle with center $(-1, 2)$ containing the point $(4, 3)$.

Background

Topic: Equation of a Circle

This question tests your ability to write the equation of a circle given its center and a point on the circle.

Key Terms and Formulas:

• Standard form: $(x - h)^2 + (y - k)^2 = r^2$

• $(h, k)$ is the center; $r$ is the radius.

• Find $r$ by calculating the distance between the center and the given point.

Step-by-Step Guidance

1. Identify the center: $(-1, 2)$

2. Calculate the radius $r$ using the distance formula between $(-1, 2)$ and $(4, 3)$: $r = \sqrt{(4 - (-1))^2 + (3 - 2)^2}$

3. Plug $h$, $k$, and $r$ into the standard circle equation.

Try solving on your own before revealing the answer!

Final Answer:

Radius: $r = \sqrt{(4 + 1)^2 + (3 - 2)^2} = \sqrt{25 + 1} = \sqrt{26}$

Equation: $(x + 1)^2 + (y - 2)^2 = 26$

Plug the center and radius into the standard form.

Q8. Give the center and radius of the circle: $(x + 2)^2 + (y - 1)^2 = 4$. Graph the circle on the grid provided.

Background

Topic: Circle Equation and Graphing

This question tests your ability to identify the center and radius from the standard form of a circle equation and graph it.

Key Terms and Formulas:

• Standard form: $(x - h)^2 + (y - k)^2 = r^2$

• Center: $(h, k)$; Radius: $r$

Step-by-Step Guidance

1. Compare the given equation to the standard form.

2. Identify $h$ and $k$ from $(x + 2)^2$ and $(y - 1)^2$.

3. Identify $r$ from $r^2 = 4$.

4. Plot the center and draw a circle with radius $r$ on the grid.

Try solving on your own before revealing the answer!

Final Answer:

Center: $(-2, 1)$

Radius: $2$

Graph: Circle centered at $(-2, 1)$ with radius $2$.

Q9. For $f(x) = 5x + 2$, $g(x) = x^2 - x$, find $f\left(g(x)\right)$ and give the domain of $f \circ g$.

Background

Topic: Function Composition and Domain

This question tests your ability to compose two functions and determine the domain of the composition.

Key Terms and Formulas:

• Composition: $f \circ g(x) = f(g(x))$

• Domain: Set of $x$ values for which $g(x)$ is defined and $f(g(x))$ is defined.

Step-by-Step Guidance

1. Find $g(x) = x^2 - x$

2. Substitute $g(x)$ into $f(x)$: $f(g(x)) = 5(g(x)) + 2$

3. Expand: $f(g(x)) = 5(x^2 - x) + 2$

4. Simplify the expression.

5. Determine the domain: Since both $f$ and $g$ are defined for all real numbers, the domain is all real numbers.

Try solving on your own before revealing the answer!

Final Answer:

$f(g(x)) = 5x^2 - 5x + 2$

Domain: All real numbers ($\mathbb{R}$)

Q10. If $f(x) = x^2 - 1$ and $g(x) = x^2 + 5$,

• (a) Find $f \circ g(x)$ and give the domain of $f \circ g$

• (b) Find $g \circ f(x)$ and give the domain of $g \circ f$

Background

Topic: Function Composition and Domain

This question tests your ability to compose functions and determine the domain of the composition.

Key Terms and Formulas:

• Composition: $f \circ g(x) = f(g(x))$ and $g \circ f(x) = g(f(x))$

• Domain: Set of $x$ values for which the composition is defined.

Step-by-Step Guidance

1. For (a), substitute $g(x)$ into $f(x)$: $f(g(x)) = (g(x))^2 - 1$

2. Expand $g(x) = x^2 + 5$ and compute $(x^2 + 5)^2 - 1$

3. For (b), substitute $f(x)$ into $g(x)$: $g(f(x)) = (f(x))^2 + 5$

4. Expand $f(x) = x^2 - 1$ and compute $(x^2 - 1)^2 + 5$

5. Both functions are defined for all real numbers, so the domain is all real numbers.

Try solving on your own before revealing the answer!

Final Answer:

(a) $f \circ g(x) = (x^2 + 5)^2 - 1$

Domain: All real numbers ($\mathbb{R}$)

(b) $g \circ f(x) = (x^2 - 1)^2 + 5$

Domain: All real numbers ($\mathbb{R}$)

Q11. Given $f(x) = |2x + 3|$,

• (a) Describe how to graph $f$ using transformations

• (b) Graph each of the functions listed in part (a) on the grids provided

Background

Topic: Absolute Value Functions and Transformations

This question tests your understanding of how to graph absolute value functions using shifts and stretches.

Key Terms and Formulas:

• Absolute value function: $f(x) = |x|$

• Transformations: Horizontal shifts, vertical shifts, and stretches/compressions.

Step-by-Step Guidance

1. Start with the parent function $y = |x|$.

2. Multiply $x$ by 2: $y = |2x|$ (horizontal compression by factor of $\frac{1}{2}$).

3. Add 3 inside the absolute value: $y = |2x + 3|$ (shift left by $\frac{3}{2}$ units).

4. Describe the sequence: compress horizontally, then shift left.

5. Graph the resulting function on the provided grid.

Try solving on your own before revealing the answer!

Final Answer:

(a) The graph of $f(x) = |2x + 3|$ is obtained by horizontally compressing $y = |x|$ by a factor of $\frac{1}{2}$ and shifting left by $\frac{3}{2}$ units.

(b) Graph: V-shaped graph with vertex at $x = -\frac{3}{2}$.

Q12. If $f(x) = x^2 - 1$ and $g(x) = x + 1$, find $f(g(x - 2))$

Background

Topic: Function Composition and Evaluation

This question tests your ability to compose functions and evaluate at a shifted argument.

Key Terms and Formulas:

• Composition: $f(g(x - 2))$

• Evaluate $g(x - 2)$ first, then substitute into $f$.

Step-by-Step Guidance

1. Compute $g(x - 2) = (x - 2) + 1 = x - 1$

2. Substitute $x - 1$ into $f$: $f(x - 1) = (x - 1)^2 - 1$

3. Expand $(x - 1)^2$ and simplify the expression.

Try solving on your own before revealing the answer!

Final Answer:

$f(g(x - 2)) = (x - 1)^2 - 1 = x^2 - 2x$

Expand and simplify to get the final expression.

Q13. If $f(x) = x^2 + 2$ and $g(x) = x^2 - 4$, find $f(g(x - 2))$

Background

Topic: Function Composition and Evaluation

This question tests your ability to compose functions and evaluate at a shifted argument.

Key Terms and Formulas:

• Composition: $f(g(x - 2))$

• Evaluate $g(x - 2)$ first, then substitute into $f$.

Step-by-Step Guidance

1. Compute $g(x - 2) = (x - 2)^2 - 4$

2. Expand $(x - 2)^2$ to $x^2 - 4x + 4$

3. So $g(x - 2) = x^2 - 4x + 4 - 4 = x^2 - 4x$

4. Substitute $x^2 - 4x$ into $f$: $f(x^2 - 4x) = (x^2 - 4x)^2 + 2$

5. Expand and simplify the expression.

Try solving on your own before revealing the answer!

Final Answer:

$f(g(x - 2)) = (x^2 - 4x)^2 + 2 = x^4 - 8x^3 + 16x^2 + 2$

Expand and combine like terms for the final expression.