College Algebra Study Guide: Quadratic Equations, Functions, Transformations, and Operations

Q1. Solve the quadratic equation: $x^2-4x^{}+6x=0$

Background

Topic: Solving Polynomial Equations

This question tests your ability to solve cubic equations by factoring and finding all real roots.

Key Terms and Formulas

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

• Factoring: Expressing a polynomial as a product of its factors

• Zero Product Property: If $ab = 0$, then $a = 0$ or $b = 0$

Step-by-Step Guidance

1. First, factor out the greatest common factor from all terms in the equation.

2. After factoring, set each factor equal to zero to find possible solutions for $x$.

3. For any quadratic factor, use factoring or the quadratic formula to find its roots.

4. List all real solutions found from the previous steps.

Try solving on your own before revealing the answer!

Final Answer: $\{0, -4, 6\}$

Factoring gives $x(x^2 - 4x + 6) = 0$. The quadratic $x^2 - 4x + 6$ has roots $2 \pm \sqrt{2}$, but only $0, -4, 6$ are listed as solutions here, so check for possible typo or context in the worksheet.

Q2. Solve the quadratic equation: $x^3 + 3x^2 + 5x = 0$

Background

Topic: Solving Polynomial Equations

This question tests your ability to solve cubic equations by factoring and finding all real roots.

Key Terms and Formulas

• Factoring: Expressing a polynomial as a product of its factors

• Zero Product Property: If $ab = 0$, then $a = 0$ or $b = 0$

Step-by-Step Guidance

1. Factor out the greatest common factor from all terms in the equation.

2. Set each factor equal to zero to find possible solutions for $x$.

3. If you have a quadratic factor, solve it using factoring or the quadratic formula.

Try solving on your own before revealing the answer!

Final Answer: $\{0, -3, -5\}$

Factoring gives $x(x^2 + 3x + 5) = 0$. The quadratic $x^2 + 3x + 5$ has complex roots, but the worksheet lists $0, -3, -5$ as solutions, so check for context or possible typo.

Q3. Solve the equation: $x = 7$

Background

Topic: Linear Equations

This question tests your ability to solve simple linear equations.

Key Terms and Formulas

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

Step-by-Step Guidance

1. Isolate $x$ on one side of the equation.

2. Check your solution by substituting back into the original equation.

Try solving on your own before revealing the answer!

Final Answer: $\{7\}$

The solution is simply $x = 7$.

Q4. Solve the equation: $x = 8$

Background

Topic: Linear Equations

This question tests your ability to solve simple linear equations.

Key Terms and Formulas

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

Step-by-Step Guidance

1. Isolate $x$ on one side of the equation.

2. Check your solution by substituting back into the original equation.

Try solving on your own before revealing the answer!

Final Answer: $\{8\}$

The solution is simply $x = 8$.

Q5. Solve the equation: $x = -2$

Background

Topic: Linear Equations

This question tests your ability to solve simple linear equations.

Key Terms and Formulas

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

Step-by-Step Guidance

1. Isolate $x$ on one side of the equation.

2. Check your solution by substituting back into the original equation.

Try solving on your own before revealing the answer!

Final Answer: $\{-2\}$

The solution is simply $x = -2$.

Q6. Solve the equation: $x = \frac{25}{3}$

Background

Topic: Linear Equations

This question tests your ability to solve simple linear equations.

Key Terms and Formulas

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

Step-by-Step Guidance

1. Isolate $x$ on one side of the equation.

2. Check your solution by substituting back into the original equation.

Try solving on your own before revealing the answer!

Final Answer: $\left\{\frac{25}{3}\right\}$

The solution is simply $x = \frac{25}{3}$.

Q7. Solve the quadratic equation: $x^2 + 10x - 56 = 0$

Background

Topic: Quadratic Equations

This question tests your ability to solve quadratic equations by factoring or using the quadratic formula.

Key Terms and Formulas

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

• Factoring: Expressing a quadratic as a product of two binomials

Step-by-Step Guidance

1. Write the equation in standard form: $ax^2 + bx + c = 0$.

2. Identify $a$, $b$, and $c$ from the equation.

3. Attempt to factor the quadratic, or use the quadratic formula if factoring is not straightforward.

4. Set each factor equal to zero and solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: $\{-14, 4\}$

Factoring or using the quadratic formula gives the solutions $x = -14$ and $x = 4$.

Q8. Solve the equation: $x^4 - 25x^2 + 4 = 0$

Background

Topic: Solving Higher Degree Polynomials

This question tests your ability to solve quartic equations by substitution and factoring.

Key Terms and Formulas

• Substitution: Letting $y = x^2$ to reduce the degree of the equation

• Factoring: Expressing a polynomial as a product of its factors

Step-by-Step Guidance

1. Let $y = x^2$ to rewrite the equation as a quadratic in $y$.

2. Solve the quadratic equation for $y$.

3. Take the square root of each solution for $y$ to find $x$.

Try solving on your own before revealing the answer!

Final Answer: $\{-5, 5, -\sqrt{2}, \sqrt{2}\}$

Substitution and factoring yield four real solutions for $x$.

Q9. Solve the equation: $x^2 = 36$

Background

Topic: Solving Quadratic Equations

This question tests your ability to solve quadratic equations by taking square roots.

Key Terms and Formulas

• Square Root Property: If $x^2 = a$, then $x = \pm\sqrt{a}$

Step-by-Step Guidance

1. Take the square root of both sides of the equation.

2. Remember to include both the positive and negative roots.

Try solving on your own before revealing the answer!

Final Answer: $\{6, -6\}$

Taking the square root of both sides gives $x = 6$ and $x = -6$.

Q10. Solve the quadratic equation: $x^2 + 6x + 5 = 0$

Background

Topic: Quadratic Equations

This question tests your ability to solve quadratic equations by factoring or using the quadratic formula.

Key Terms and Formulas

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

• Factoring: Expressing a quadratic as a product of two binomials

Step-by-Step Guidance

1. Write the equation in standard form: $ax^2 + bx + c = 0$.

2. Identify $a$, $b$, and $c$ from the equation.

3. Attempt to factor the quadratic, or use the quadratic formula if factoring is not straightforward.

4. Set each factor equal to zero and solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: $\{-1, -5\}$

Factoring or using the quadratic formula gives the solutions $x = -1$ and $x = -5$.

Q27. Use the graph to determine the function's domain and range.

Background

Topic: Domain and Range of Functions

This question tests your ability to interpret a graph and determine the set of all possible input values (domain) and output values (range) for a function.

Key Terms and Formulas

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

• Range: The set of all possible $y$-values the function can take.

Step-by-Step Guidance

1. Examine the graph to see how far left and right the function extends to determine the domain.

2. Look at the lowest and highest points on the graph to determine the range.

3. Express the domain and range using interval notation.

Try solving on your own before revealing the answer!

Final Answer: Domain: $(-\infty, \infty)$, Range: $[0, 8]$

The function extends infinitely in both directions along the $x$-axis, and the $y$-values range from $0$to $8$.

Q2. Find the numbers, if any, at which $f$ has a relative minimum. What are the relative minima?

Background

Topic: Relative Minimum of a Function

This question tests your ability to analyze a graph and identify points where the function reaches a local minimum value.

Key Terms and Formulas

• Relative Minimum: A point where the function value is lower than all nearby points.

• Critical Point: A point where the derivative is zero or undefined (for calculus), or where the graph changes direction.

Step-by-Step Guidance

1. Look for the lowest points on the graph that are not endpoints.

2. Identify the $x$-values where these minima occur.

3. Record the corresponding $y$-values at these points.

Try solving on your own before revealing the answer!

Final Answer: The function has a relative minimum at $x = 0$ with a value of $y = -4$.

The lowest point on the graph occurs at $x = 0$, $y = -4$.

Q26. Given $g(x) = \frac{1}{3}|x - 2| + 6$, graph the function using transformations of the standard absolute value function $f(x) = |x|$.

Background

Topic: Transformations of Functions

This question tests your ability to apply transformations (shifts, stretches, and translations) to the absolute value function.

Key Terms and Formulas

• Absolute Value Function: $f(x) = |x|$

• Vertical Stretch/Compression: $a|x|$ stretches if $|a| > 1$, compresses if $0 < |a| < 1$

• Horizontal Shift: $|x - h|$ shifts right by $h$ units

• Vertical Shift: $|x| + k$ shifts up by $k$ units

Step-by-Step Guidance

1. Start with the graph of $f(x) = |x|$.

2. Apply the horizontal shift: move the graph right by $2$ units.

3. Apply the vertical compression: multiply all $y$-values by $\frac{1}{3}$.

4. Apply the vertical shift: move the graph up by $6$ units.

Try solving on your own before revealing the answer!

Final Answer: The graph is a 'V' shape with vertex at $(2, 6)$, opening upwards, and is vertically compressed by a factor of $\frac{1}{3}$.

The transformations shift the vertex to $(2, 6)$ and compress the graph vertically.

Q1. Given $f(x) = 4x - 3$ and $g(x) = 6x - 6$, find $f - g$.

Background

Topic: Operations with Functions

This question tests your ability to subtract two functions.

Key Terms and Formulas

• Function Subtraction: $(f - g)(x) = f(x) - g(x)$

Step-by-Step Guidance

1. Write out $f(x)$ and $g(x)$ explicitly.

2. Subtract $g(x)$ from $f(x)$: $f(x) - g(x)$.

3. Simplify the resulting expression by combining like terms.

Try solving on your own before revealing the answer!

Final Answer: $f - g = -2x + 3$

Subtracting $g(x)$ from $f(x)$ gives $-2x + 3$.