College Algebra Exam 2 Review: Step-by-Step Guidance

Q1. Find the zeros of the following functions and state each zero as an ordered pair. Graph the function using intercepts, axis of symmetry, and vertex. Label all points and draw the axis of symmetry as a dotted line.

• (a) $f(x) = 6x^2 + 13x - 5$

• (b) $g(x) = x^2 + x - 6$

• (c) $h(x) = 3x^2 - 12$

• (d) $f(x) = (x - 2)^2 - 16$

• (e) $g(x) = 4x^2 - x + 2$

• (f) $h(x) = 3x^2 - 5x + 1$

Background

Topic: Quadratic Functions and Graphing

This question tests your ability to find the zeros (roots) of quadratic functions, express them as ordered pairs, and graph the functions by identifying intercepts, the axis of symmetry, and the vertex.

Key Terms and Formulas

• Zero of a function: The value(s) of $x$ where $f(x) = 0$.

• Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for $ax^2 + bx + c = 0$.

• Vertex: $x = -\frac{b}{2a}$ for $f(x) = ax^2 + bx + c$.

• Axis of symmetry: $x = -\frac{b}{2a}$.

• Y-intercept: Set $x = 0$ and solve for $f(0)$.

Step-by-Step Guidance

1. For each function, set $f(x) = 0$ and solve for $x$ using factoring, completing the square, or the quadratic formula as appropriate.

2. Express each zero as an ordered pair $(x, 0)$.

3. Find the y-intercept by evaluating $f(0)$ and write it as $(0, f(0))$.

4. Determine the axis of symmetry using $x = -\frac{b}{2a}$ (or $x = h$ if in vertex form).

5. Find the vertex by plugging the axis of symmetry value into the function: $(h, f(h))$.

6. Set up your graph: plot the zeros, y-intercept, vertex, and draw the axis of symmetry as a dotted line. Label all points.

Try solving on your own before revealing the answer!

Final Answer:

Each function's zeros, vertex, y-intercept, and axis of symmetry are as follows:

• (a) $f(x) = 6x^2 + 13x - 5$ Zeros: $x = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 6 \cdot (-5)}}{2 \cdot 6} = \frac{-13 \pm \sqrt{169 + 120}}{12} = \frac{-13 \pm \sqrt{289}}{12} = \frac{-13 \pm 17}{12}$ So, $x_1 = \frac{4}{12} = \frac{1}{3}$, $x_2 = \frac{-30}{12} = -\frac{5}{2}$ Ordered pairs: $(\frac{1}{3}, 0)$ and $(-\frac{5}{2}, 0)$ Vertex: $x = -\frac{13}{2 \cdot 6} = -\frac{13}{12}$, $f(-\frac{13}{12}) = 6(-\frac{13}{12})^2 + 13(-\frac{13}{12}) - 5 \approx -10.04$ Vertex: $(-\frac{13}{12}, -10.04)$ Y-intercept: $f(0) = -5$ at $(0, -5)$ Axis of symmetry: $x = -\frac{13}{12}$

• (b) $g(x) = x^2 + x - 6$ Zeros: $x = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}$ So, $x_1 = 2$, $x_2 = -3$ Ordered pairs: $(2, 0)$ and $(-3, 0)$ Vertex: $x = -\frac{1}{2}$, $g(-\frac{1}{2}) = (-\frac{1}{2})^2 + (-\frac{1}{2}) - 6 = -6.25$ Vertex: $(-0.5, -6.25)$ Y-intercept: $g(0) = -6$ at $(0, -6)$ Axis of symmetry: $x = -0.5$

• (c) $h(x) = 3x^2 - 12$ Zeros: $3x^2 - 12 = 0 \implies x^2 = 4 \implies x = \pm 2$ Ordered pairs: $(2, 0)$ and $(-2, 0)$ Vertex: $x = 0$, $h(0) = -12$ Vertex: $(0, -12)$ Y-intercept: $(0, -12)$ Axis of symmetry: $x = 0$

• (d) $f(x) = (x - 2)^2 - 16$ Zeros: $(x - 2)^2 = 16 \implies x - 2 = \pm 4 \implies x = 6$ or $x = -2$ Ordered pairs: $(6, 0)$ and $(-2, 0)$ Vertex: $(2, -16)$ Y-intercept: $f(0) = (0 - 2)^2 - 16 = 4 - 16 = -12$ at $(0, -12)$ Axis of symmetry: $x = 2$

• (e) $g(x) = 4x^2 - x + 2$ Zeros: $x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 4 \cdot 2}}{2 \cdot 4} = \frac{1 \pm \sqrt{1 - 32}}{8}$ Discriminant is negative, so no real zeros. Vertex: $x = \frac{1}{8}$, $g(\frac{1}{8}) = 4(\frac{1}{8})^2 - (\frac{1}{8}) + 2 \approx 1.89$ Vertex: $(0.125, 1.89)$ Y-intercept: $g(0) = 2$ at $(0, 2)$ Axis of symmetry: $x = 0.125$

• (f) $h(x) = 3x^2 - 5x + 1$ Zeros: $x = \frac{5 \pm \sqrt{25 - 12}}{6} = \frac{5 \pm \sqrt{13}}{6}$ Ordered pairs: $\left(\frac{5 + \sqrt{13}}{6}, 0\right)$ and $\left(\frac{5 - \sqrt{13}}{6}, 0\right)$ Vertex: $x = \frac{5}{6}$, $h(\frac{5}{6}) = 3(\frac{5}{6})^2 - 5(\frac{5}{6}) + 1 \approx -1.08$ Vertex: $(0.83, -1.08)$ Y-intercept: $h(0) = 1$ at $(0, 1)$ Axis of symmetry: $x = 0.83$

For each graph, plot the zeros (if real), y-intercept, vertex, and draw the axis of symmetry as a dotted line. Label all points as required.

Q2. Solve the following equation. Remember to check your solution(s):

$2 + \frac{12}{-2x} = x$

Background

Topic: Rational Equations

This question tests your ability to solve equations involving rational expressions and to check for extraneous solutions.

Key Terms and Formulas

• Rational equation: An equation involving fractions whose numerators and/or denominators contain a variable.

• Extraneous solution: A solution that does not satisfy the original equation, often due to division by zero.

Step-by-Step Guidance

1. Identify the denominators and determine any restrictions on $x$ (values that would make a denominator zero).

2. Multiply both sides of the equation by the least common denominator (LCD) to eliminate fractions.

3. Simplify the resulting equation and solve for $x$.

4. Check your solution(s) in the original equation to ensure they do not make any denominator zero.

Try solving on your own before revealing the answer!

Final Answer:

Solving $2 + \frac{12}{-2x} = x$ gives $x = 1$ or $x = -6$. Both values are valid since they do not make the denominator zero.

Q3. State the solution set of each inequality in inequality or interval notation:

• (a) $0 < 3x + 2$

• (b) $\frac{2}{4} < \frac{1}{2} \leq x + 1 \leq \frac{3}{4}$

• (c) $x^2 - 1 < 0$

• (d) $x^2 - 2x - 24 \geq 0$

Background

Topic: Solving Linear and Quadratic Inequalities

This question tests your ability to solve inequalities and express the solution set in both inequality and interval notation.

Key Terms and Formulas

• Linear inequality: An inequality involving a linear expression.

• Quadratic inequality: An inequality involving a quadratic expression.

• Interval notation: A way of writing subsets of the real number line.

Step-by-Step Guidance

1. For each inequality, isolate the variable using algebraic operations.

2. For quadratic inequalities, factor the expression if possible and determine the intervals where the inequality holds by testing values or using a sign chart.

3. Express the solution set in both inequality and interval notation.

4. For compound inequalities, break them into parts and solve each part, then combine the results.

Try solving on your own before revealing the answer!

Final Answer:

• (a) $x > -\frac{2}{3}$; interval: $(-\frac{2}{3}, \infty)$

• (b) $0.5 < x + 1 \leq 0.75 \implies -0.5 < x \leq -0.25$; interval: $(-0.5, -0.25]$

• (c) $-1 < x < 1$; interval: $(-1, 1)$

• (d) $x \leq -4$ or $x \geq 6$; interval: $(-\infty, -4] \cup [6, \infty)$

Q4. Solve the following absolute value equation. Remember to check your solution(s):

$|3x - 1| = 2$

Background

Topic: Absolute Value Equations

This question tests your ability to solve equations involving absolute values and to check for extraneous solutions.

Key Terms and Formulas

• Absolute value equation: $|A| = B$ implies $A = B$ or $A = -B$.

Step-by-Step Guidance

1. Set up two equations: $3x - 1 = 2$ and $3x - 1 = -2$.

2. Solve each equation for $x$.

3. Check each solution in the original equation to ensure it is valid.

Try solving on your own before revealing the answer!

Final Answer:

Solutions: $x = 1$ and $x = -\frac{1}{3}$

Q5. State the solution set of each absolute value inequality in inequality or interval notation:

• (a) $|1 - 2x| - 4 < -1$

• (b) $|3x + 4| \geq 2$

Background

Topic: Absolute Value Inequalities

This question tests your ability to solve inequalities involving absolute values and express the solution set in interval notation.

Key Terms and Formulas

• For $|A| < B$, solution is $-B < A < B$.

• For $|A| \geq B$, solution is $A \leq -B$ or $A \geq B$.

Step-by-Step Guidance

1. For each inequality, isolate the absolute value expression.

2. Rewrite the inequality as two separate inequalities (as appropriate for $<$ or $\geq$).

3. Solve each inequality for $x$.

4. Express the solution set in interval notation.

Try solving on your own before revealing the answer!

Final Answer:

• (a) $|1 - 2x| < 3$ so $-3 < 1 - 2x < 3$; solution: $-1 < x < 2$; interval: $(-1, 2)$

• (b) $|3x + 4| \geq 2$ so $3x + 4 \leq -2$ or $3x + 4 \geq 2$; solution: $x \leq -2$ or $x \geq -\frac{2}{3}$; interval: $(-\infty, -2] \cup [-\frac{2}{3}, \infty)$

Q6. Find the equation of the quadratic function that passes through a vertex of (2, -3) and the point (0, 7). You may leave it in vertex form.

Background

Topic: Quadratic Functions (Vertex Form)

This question tests your ability to write the equation of a quadratic function given its vertex and another point.

Key Terms and Formulas

• Vertex form: $f(x) = a(x - h)^2 + k$, where $(h, k)$ is the vertex.

Step-by-Step Guidance

1. Write the general vertex form using the given vertex $(2, -3)$: $f(x) = a(x - 2)^2 - 3$.

2. Substitute the coordinates of the given point $(0, 7)$ into the equation to solve for $a$.

3. Solve for $a$ and write the final equation in vertex form.

Try solving on your own before revealing the answer!

Final Answer:

Substitute $(0, 7)$ into $f(x) = a(x - 2)^2 - 3$:

$7 = a(0 - 2)^2 - 3 \implies 7 = 4a - 3 \implies a = 2.5$

So, the equation is $f(x) = 2.5(x - 2)^2 - 3$

Q7. Given the quadratic function $f(x) = -3x^2 + 6x + 1$, use algebra to find:

• (a) Is $f(x)$ concave up or concave down?

• (b) The y-intercept as an ordered pair

• (c) The equation for the axis of symmetry

• (d) The vertex as an ordered pair

• (e) All x-intercepts as ordered pairs

Background

Topic: Properties of Quadratic Functions

This question tests your understanding of the shape, intercepts, axis of symmetry, and vertex of a quadratic function.

Key Terms and Formulas

• Concavity: Determined by the sign of $a$ in $ax^2 + bx + c$.

• Y-intercept: $f(0)$

• Axis of symmetry: $x = -\frac{b}{2a}$

• Vertex: $(h, f(h))$ where $h = -\frac{b}{2a}$

• X-intercepts: Solve $f(x) = 0$

Step-by-Step Guidance

1. Identify the value of $a$ to determine concavity.

2. Find the y-intercept by evaluating $f(0)$.

3. Calculate the axis of symmetry using $x = -\frac{b}{2a}$.

4. Find the vertex by substituting the axis of symmetry value into $f(x)$.

5. Solve $f(x) = 0$ using the quadratic formula to find x-intercepts.

Try solving on your own before revealing the answer!

Final Answer:

• (a) Concave down (since $a = -3 < 0$)

• (b) Y-intercept: $(0, 1)$

• (c) Axis of symmetry: $x = 1$

• (d) Vertex: $(1, 4)$

• (e) X-intercepts: $x = 3$ and $x = -\frac{1}{3}$; ordered pairs: $(3, 0)$ and $(-\frac{1}{3}, 0)$

Q8. The price $p$ (in dollars) and the quantity $x$ sold of a certain product satisfy the demand equation $x = -6p + 600$

• (a) Find a model that expresses the revenue $R$ as a function of $p$. (Recall, $R = xp$)

• (b) What price $p$ maximizes the revenue? What is the maximum revenue?

• (c) What price should the company charge to earn at least $12,600$ in revenue?

Background

Topic: Quadratic Revenue Models

This question tests your ability to create and analyze a quadratic revenue function based on a linear demand equation.

Key Terms and Formulas

• Revenue: $R = xp$

• Demand equation: $x = -6p + 600$

• Quadratic function: $R(p) = -6p^2 + 600p$

• Vertex of a parabola: $p = -\frac{b}{2a}$ for $R(p) = ap^2 + bp + c$

Step-by-Step Guidance

1. Substitute the demand equation into the revenue formula to get $R$ as a function of $p$.

2. Write the quadratic revenue function in standard form.

3. Find the price $p$ that maximizes revenue using the vertex formula.

4. Calculate the maximum revenue by substituting this $p$ value into $R(p)$.

5. Set $R(p) \geq 12,600$ and solve the resulting quadratic inequality for $p$.

Try solving on your own before revealing the answer!

Final Answer:

• (a) $R(p) = -6p^2 + 600p$

• (b) Maximum at $p = 50$, $R(50) = 15,000$

• (c) $R(p) \geq 12,600$ gives $p \approx 24.5$ or $p \approx 75.5$; so, the company should charge between $\$24.50$ and $\$75.50$ (rounded to the nearest cent)

Q9. An astronaut on the moon throws a baseball upward. The height $s$, in feet, of the ball $t$ seconds after the ball was thrown is given by the equation $s = -2.7t^2 + 30t + 6.5$.

• (a) After how many seconds is the ball 22 feet above the moon’s surface? State your solution(s) as a decimal rounded to the nearest hundredth.

• (b) When will the ball be at least 34 feet above the moon’s surface? State your solution as a compound inequality with values rounded to the nearest hundredth.

Background

Topic: Quadratic Equations in Physics (Projectile Motion)

This question tests your ability to solve quadratic equations and inequalities in the context of projectile motion.

Key Terms and Formulas

• Quadratic equation: $s = -2.7t^2 + 30t + 6.5$

• To find when $s = h$, set $-2.7t^2 + 30t + 6.5 = h$ and solve for $t$.

• To find when $s \geq h$, solve $-2.7t^2 + 30t + 6.5 \geq h$.

Step-by-Step Guidance

1. For (a), set $s = 22$ and write the equation: $-2.7t^2 + 30t + 6.5 = 22$.

2. Rearrange to standard quadratic form: $-2.7t^2 + 30t - 15.5 = 0$.

3. Use the quadratic formula to solve for $t$.

4. For (b), set up the inequality $-2.7t^2 + 30t + 6.5 \geq 34$.

5. Rearrange to $-2.7t^2 + 30t - 27.5 \geq 0$ and solve the corresponding quadratic equation for $t$ to find the interval.

Try solving on your own before revealing the answer!

Final Answer:

• (a) $t \approx 0.55$ seconds and $t \approx 10.45$ seconds

• (b) The ball is at least 34 feet above the surface for $0.97 \leq t \leq 10.53$ seconds