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

Q1. Use your graphing utility to approximate the intercepts rounded to three decimal places for the equation $y = x^3 - 2x^2 + 3x$.

Background

Topic: Polynomial Functions and Graphs

This question tests your ability to find the x- and y-intercepts of a cubic polynomial using a graphing calculator or graphing software.

Key Terms and Formulas:

• x-intercept: The value(s) of $x$ where $y = 0$.

• y-intercept: The value of $y$ when $x = 0$.

Step-by-Step Guidance

1. To find the y-intercept, substitute $x = 0$ into the equation: $y = 0^3 - 2 \times 0^2 + 3 \times 0$.

2. To find the x-intercepts, set $y = 0$ and solve $x^3 - 2x^2 + 3x = 0$.

3. Factor the equation if possible to make solving easier.

4. Use your graphing utility to plot the function and identify the points where the graph crosses the x-axis (these are the x-intercepts).

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Q2. Solve the equation graphically: $4x^2 - 3x^2 = 2x + 4$ (Round your solution(s) to two decimal places).

Background

Topic: Solving Equations Graphically

This question asks you to use a graphing calculator or software to find the solution(s) to a quadratic equation by identifying where two functions intersect.

Key Terms and Formulas:

• Graphical Solution: The x-values where the graphs of $y_1 = 4x^2 - 3x^2$ and $y_2 = 2x + 4$ intersect.

Step-by-Step Guidance

1. Simplify the equation: $4x^2 - 3x^2 = x^2$ so the equation becomes $x^2 = 2x + 4$.

2. Rewrite as $x^2 - 2x - 4 = 0$ to set up for graphing.

3. Graph $y = x^2 - 2x - 4$ and look for the x-values where $y = 0$ (the x-intercepts).

4. Use your graphing utility to approximate the x-values to two decimal places.

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Q3a. Solve the equation algebraically: $4(x^2) - 3(5x^2) - 6x^2 = x^2 + 3$

Background

Topic: Solving Quadratic Equations

This question tests your ability to combine like terms and solve a quadratic equation algebraically.

Key Terms and Formulas:

• Quadratic Equation: An equation of the form $ax^2 + bx + c = 0$.

• Combining Like Terms: Add or subtract terms with the same variable and exponent.

Step-by-Step Guidance

1. Expand and simplify each term: $4(x^2) = 4x^2$, $-3(5x^2) = -15x^2$, $-6x^2$ remains as is.

2. Combine all $x^2$ terms on the left side.

3. Move all terms to one side to set the equation equal to zero.

4. Factor or use the quadratic formula to solve for $x$.

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Q4. A car dealer reduces the list price of its vehicles by 14%. If one vehicle has a sale price of $28,809.99, what was its original list price to the nearest cent?

Background

Topic: Percentages and Reverse Calculations

This question tests your ability to work with percentages and solve for the original value before a reduction.

Key Terms and Formulas:

• Sale Price: The price after a percentage reduction.

• Original Price Formula: $\text{Sale Price} = \text{Original Price} \times (1 - \text{Percent Reduction})$

Step-by-Step Guidance

1. Let $P$ be the original price. The sale price is $P \times (1 - 0.14)$.

2. Set up the equation: $28,809.99 = P \times 0.86$.

3. Rearrange to solve for $P$ by dividing both sides by $0.86$.

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Q5a. If a rock falls from a height of 20 meters, the height $s$ in meters after $t$ seconds is given by $s = 20 - 4.9t^2$. What is the height of the rock after 1.5 seconds?

Background

Topic: Quadratic Functions and Applications

This question tests your ability to substitute values into a quadratic function modeling free fall.

Key Terms and Formulas:

• Free Fall Formula: $s = s_0 - 4.9t^2$ where $s_0$ is the initial height.

Step-by-Step Guidance

1. Substitute $t = 1.5$ into the formula: $s = 20 - 4.9 \times (1.5)^2$.

2. Calculate $(1.5)^2$ first.

3. Multiply the result by $4.9$.

4. Subtract this value from $20$ to find the height.

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Q5b. When does the rock strike the ground? (approximate to two decimal places)

Background

Topic: Quadratic Equations and Applications

This question tests your ability to solve for the time when the height is zero (the rock hits the ground).

Key Terms and Formulas:

• Free Fall Formula: $s = 20 - 4.9t^2$

Step-by-Step Guidance

1. Set $s = 0$ to find when the rock hits the ground: $0 = 20 - 4.9t^2$.

2. Rearrange to solve for $t^2$: $4.9t^2 = 20$.

3. Divide both sides by $4.9$ to isolate $t^2$.

4. Take the square root of both sides to solve for $t$.

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Q6a. Solve for $x$: $3x^2 - 2x^2 = 3x + 5$

Background

Topic: Quadratic Equations

This question tests your ability to simplify and solve a quadratic equation.

Key Terms and Formulas:

• Quadratic Equation: $ax^2 + bx + c = 0$

• Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Step-by-Step Guidance

1. Simplify the left side: $3x^2 - 2x^2 = x^2$.

2. Rewrite the equation: $x^2 = 3x + 5$.

3. Move all terms to one side: $x^2 - 3x - 5 = 0$.

4. Identify $a = 1$, $b = -3$, $c = -5$ for the quadratic formula.

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Q7a. Solve the equation algebraically: $4x - 3 = 0$

Background

Topic: Linear Equations

This question tests your ability to solve a simple linear equation for $x$.

Key Terms and Formulas:

• Linear Equation: An equation of the form $ax + b = 0$.

Step-by-Step Guidance

1. Add $3$to both sides: $4x = 3$.

2. Divide both sides by $4$to solve for$x$.

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Q8a. Solve the inequalities: $3x - 3 \leq 2x - 9$

Background

Topic: Solving Linear Inequalities

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

Key Terms and Formulas:

• Linear Inequality: An inequality involving linear expressions.

• Interval Notation: A way to express the solution set of an inequality.

Step-by-Step Guidance

1. Subtract $2x$ from both sides: $3x - 2x - 3 \leq -9$.

2. Simplify: $x - 3 \leq -9$.

3. Add $3$to both sides:$x \leq -6$.

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Q9a. Solve the system of equations: $\begin{cases} 2x + 5y = 3 \\ 3x - y = 9 \end{cases}$

Background

Topic: Systems of Linear Equations

This question tests your ability to solve a system of two linear equations using substitution or elimination.

Key Terms and Formulas:

• System of Equations: Two or more equations with the same variables.

• Substitution/Elimination: Methods for solving systems.

Step-by-Step Guidance

1. Solve one equation for one variable (e.g., from the second equation, $y = 3x - 9$).

2. Substitute this expression for $y$ into the first equation.

3. Solve for $x$.

4. Once you have $x$, substitute back to find $y$.

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Q10. If the area of a rectangle is 108 square inches and the perimeter is 42 inches, what are the dimensions of the rectangle?

Background

Topic: Geometry – Area and Perimeter

This question tests your ability to set up and solve a system of equations involving area and perimeter.

Key Terms and Formulas:

• Area: $A = l \times w$

• Perimeter: $P = 2l + 2w$

Step-by-Step Guidance

1. Let $l$ = length and $w$ = width.

2. Set up the equations: $l \times w = 108$ and $2l + 2w = 42$.

3. Solve one equation for one variable (e.g., $l = 108/w$).

4. Substitute into the perimeter equation and solve for $w$.

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Q11. A motorboat travels on a river that has a current of 3 mph. The trip upstream takes 5 hours and the return trip downstream takes 2.5 hours. What is the speed of the boat?

Background

Topic: Distance, Rate, and Time Problems

This question tests your ability to set up and solve equations involving rates in opposite directions with a current.

Key Terms and Formulas:

• Distance Formula: $\text{Distance} = \text{Rate} \times \text{Time}$

• Upstream Rate: $r - 3$

• Downstream Rate: $r + 3$

Step-by-Step Guidance

1. Let $r$ be the speed of the boat in still water.

2. Set up the equations for upstream and downstream: $\text{Distance} = (r - 3) \times 5$ and $\text{Distance} = (r + 3) \times 2.5$.

3. Since the distances are equal, set the two expressions equal to each other.

4. Solve for $r$.

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Q12. Donna invested $33,000 and received a total of $970 in interest after one year. If part of the money returned 4% interest and the rest returned 2.25% interest, how much did she invest at each rate?

Background

Topic: Systems of Equations – Investment Problems

This question tests your ability to set up and solve a system of equations involving investments at different rates.

Key Terms and Formulas:

• Interest Formula: $I = P \times r$

Step-by-Step Guidance

1. Let $x$ be the amount invested at 4%, and $y$ be the amount invested at 2.25%.

2. Set up the equations: $x + y = 33,000$ and $0.04x + 0.0225y = 970$.

3. Solve one equation for one variable and substitute into the other.

4. Solve for $x$ and $y$.

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Q13. How much water should be added to 300 milliliters of a 60% acid solution to make a 50% acid solution?

Background

Topic: Mixture Problems

This question tests your ability to set up and solve a mixture equation.

Key Terms and Formulas:

• Mixture Formula: $\text{Amount of acid} = \text{Total volume} \times \text{Concentration}$

Step-by-Step Guidance

1. Let $x$ be the amount of water to add.

2. The new total volume is $300 + x$.

3. The amount of acid remains $0.6 \times 300$.

4. Set up the equation: $0.6 \times 300 = 0.5 \times (300 + x)$.

5. Solve for $x$.

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Q14. An 8% solution and a 15% solution are to be mixed to obtain 20 ounces of a 10% solution. How much of each should be used?

Background

Topic: Mixture Problems

This question tests your ability to set up and solve a system of equations for mixture concentrations.

Key Terms and Formulas:

• Mixture Formula: $\text{Total acid} = \text{Volume}_1 \times \text{Concentration}_1 + \text{Volume}_2 \times \text{Concentration}_2$

Step-by-Step Guidance

1. Let $x$ be the amount of 8% solution, $y$ be the amount of 15% solution.

2. Set up the equations: $x + y = 20$ and $0.08x + 0.15y = 0.10 \times 20$.

3. Solve one equation for one variable and substitute into the other.

4. Solve for $x$ and $y$.

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Q15. Two cyclists leave a city at the same time, one going east and the other going west. The westbound cyclist bikes 5 mph faster than the eastbound cyclist. After 6 hours they are 246 miles apart. How fast is each cyclist riding?

Background

Topic: Distance, Rate, and Time Problems

This question tests your ability to set up and solve equations involving rates and total distance.

Key Terms and Formulas:

• Distance Formula: $\text{Distance} = \text{Rate} \times \text{Time}$

Step-by-Step Guidance

1. Let $r$ be the speed of the eastbound cyclist, $r + 5$ for the westbound.

2. Each travels for 6 hours: $6r$ and $6(r + 5)$.

3. Total distance apart: $6r + 6(r + 5) = 246$.

4. Solve for $r$.

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Q16c. Find an equation of the line parallel to $3x + 4y = 2$ and through the point (4, 1).

Background

Topic: Linear Equations – Parallel Lines

This question tests your ability to find the equation of a line parallel to a given line and passing through a specific point.

Key Terms and Formulas:

• Slope-Intercept Form: $y = mx + b$

• Parallel Lines: Have the same slope.

Step-by-Step Guidance

1. Rewrite $3x + 4y = 2$ in slope-intercept form to find the slope.

2. Use the same slope for your new line.

3. Plug the point (4, 1) into the equation to solve for $b$.

4. Write the final equation in slope-intercept form.

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Q17. Find the intercepts of the line $5x - 2y = 8$. Use the intercepts to graph the line.

Background

Topic: Linear Equations – Intercepts

This question tests your ability to find x- and y-intercepts and use them to graph a line.

Key Terms and Formulas:

• x-intercept: Set $y = 0$ and solve for $x$.

• y-intercept: Set $x = 0$ and solve for $y$.

Step-by-Step Guidance

1. Set $y = 0$ in $5x - 2y = 8$ and solve for $x$.

2. Set $x = 0$ in $5x - 2y = 8$ and solve for $y$.

3. Plot these points on a graph and draw the line through them.

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Q18. An electric company charges a flat monthly fee of $12.76 plus an additional $0.18$ for every kilowatt-hour of usage. Write a linear equation that gives the monthly charge $C$ for a customer who uses $x$ kilowatt-hours in a month.

Background

Topic: Linear Functions – Applications

This question tests your ability to write a linear equation from a real-world scenario.

Key Terms and Formulas:

• Linear Equation: $C = mx + b$

Step-by-Step Guidance

1. Identify the fixed fee ($b = 12.76$) and the variable rate ($m = 0.18$).

2. Write the equation: $C = 0.18x + 12.76$.

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Q19. The volume of a gas in a closed container varies inversely with its pressure. If the volume is 600 cm³ when the pressure is 150 mm Hg, find the volume when the pressure is 200 mm Hg.

Background

Topic: Inverse Variation

This question tests your ability to apply the concept of inverse variation to solve for an unknown.

Key Terms and Formulas:

• Inverse Variation: $V = \frac{k}{P}$

Step-by-Step Guidance

1. Use the given values to find the constant $k$: $600 = \frac{k}{150}$.

2. Solve for $k$.

3. Use $k$ to find the new volume when $P = 200$.

4. Set up $V = \frac{k}{200}$.

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Q20. The resistance of a wire varies directly with its length and inversely with the square of the diameter. If a wire 1.2 meters long and 0.5 cm in diameter has a resistance of 140 ohms, find the resistance of a wire made from the same material that is 3 meters long and has a diameter of 0.8 cm.

Background

Topic: Direct and Inverse Variation

This question tests your ability to set up and solve a variation equation.

Key Terms and Formulas:

• Variation Formula: $R = k \frac{L}{d^2}$

Step-by-Step Guidance

1. Use the first set of values to solve for $k$: $140 = k \frac{1.2}{(0.5)^2}$.

2. Solve for $k$.

3. Use $k$ to find the resistance for the new wire: $R = k \frac{3}{(0.8)^2}$.

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Q21a. Find and simplify $f(3)$ for $f(x) = 2x^3 - x + 3$.

Background

Topic: Function Evaluation

This question tests your ability to substitute a value into a function and simplify.

Key Terms and Formulas:

• Function Evaluation: Substitute $x$ with the given value.

Step-by-Step Guidance

1. Substitute $x = 3$ into $f(x)$: $f(3) = 2 \times 3^3 - 3 + 3$.

2. Calculate $3^3$ first.

3. Multiply by $2$.

4. Subtract $3$and add $3$.

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Q22a. Find the domain of $f(x) = \frac{2x^3 + 4}{12 - x}$.

Background

Topic: Domain of Rational Functions

This question tests your ability to find the domain of a function with a denominator.

Key Terms and Formulas:

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

• Rational Function: A function of the form $\frac{P(x)}{Q(x)}$.

Step-by-Step Guidance

1. Identify values of $x$ that make the denominator zero: $12 - x = 0$.

2. Solve for $x$.

3. The domain excludes this value.

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Q24a. If a rock falls from a height of 25 meters, the height $s$ in meters after $t$ seconds is given by $s(t) = 25 - 4.9t^2$. What is the height after 1.2 seconds?

Background

Topic: Quadratic Functions – Applications

This question tests your ability to substitute a value into a quadratic function modeling free fall.

Key Terms and Formulas:

• Free Fall Formula: $s(t) = s_0 - 4.9t^2$

Step-by-Step Guidance

1. Substitute $t = 1.2$ into the formula: $s = 25 - 4.9 \times (1.2)^2$.

2. Calculate $(1.2)^2$ first.

3. Multiply by $4.9$.

4. Subtract this value from $25$ to find the height.

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