College Algebra Review: Quadratic, Polynomial, Rational, Exponential, and Logarithmic Functions

Q1. Let $f(x) = -2x^2 + 8x$.

Background

Topic: Quadratic Functions

This question tests your understanding of the properties of quadratic functions, including graph orientation, vertex, axis of symmetry, and intercepts.

Key Terms and Formulas:

• Standard form of a quadratic: $f(x) = ax^2 + bx + c$

• Vertex: $\left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right)$

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

• $x$-intercepts: Set $f(x) = 0$ and solve for $x$

• $y$-intercept: $f(0)$

Step-by-Step Guidance

1. Identify the leading coefficient $a$ in $f(x) = -2x^2 + 8x$. Determine if the parabola opens upward or downward based on the sign of $a$.

2. Find the vertex using the formula $x = -\frac{b}{2a}$. Substitute $a = -2$ and $b = 8$ to find the $x$-coordinate of the vertex.

3. Plug the $x$-coordinate of the vertex back into $f(x)$ to find the $y$-coordinate of the vertex.

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

5. To find the $x$-intercepts, set $f(x) = 0$ and solve the quadratic equation $-2x^2 + 8x = 0$ for $x$.

6. To find the $y$-intercept, substitute $x = 0$ into $f(x)$ and simplify.

Try solving on your own before revealing the answer!

Final Answer:

(a) The graph opens downward because $a = -2 < 0$.

(b) Vertex: $(2, 8)$; Axis of symmetry: $x = 2$.

(c) $x$-intercepts: $(0, 0)$ and $(4, 0)$.

(d) $y$-intercept: $(0, 0)$.

We used the vertex formula and solved for intercepts as shown in the steps above.

Q2. Consider the quadratic function $f(x) = 3x^2 + 6x - 9$.

Background

Topic: Quadratic Functions (Vertex Form, Domain/Range, Intercepts)

This question tests your ability to analyze a quadratic function, including graph orientation, vertex, axis of symmetry, vertex form, domain, range, and intercepts.

Key Terms and Formulas:

• Standard form: $f(x) = ax^2 + bx + c$

• Vertex: $\left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right)$

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

• Vertex form: $f(x) = a(x-h)^2 + k$

• Domain of any quadratic: $(-\infty, \infty)$

• Range: Depends on the vertex and whether the parabola opens up or down

• Intercepts: Set $f(x) = 0$ for $x$-intercepts, $f(0)$ for $y$-intercept

Step-by-Step Guidance

1. Identify $a$, $b$, and $c$ in $f(x) = 3x^2 + 6x - 9$. Use the sign of $a$ to determine if the parabola opens upward or downward.

2. Find the vertex using $x = -\frac{b}{2a}$, then substitute back into $f(x)$ to get the $y$-coordinate.

3. Write $f(x)$ in vertex form by completing the square.

4. State the domain (all real numbers) and determine the range based on the vertex and direction of opening.

5. Find the $x$-intercepts by solving $3x^2 + 6x - 9 = 0$ and the $y$-intercept by evaluating $f(0)$.

Try solving on your own before revealing the answer!

Final Answer:

(a) Opens upward ($a = 3 > 0$).

(b) Vertex: $(-1, -12)$; Axis of symmetry: $x = -1$.

(c) Vertex form: $f(x) = 3(x+1)^2 - 12$.

(d) Domain: $(-\infty, \infty)$; Range: $[-12, \infty)$.

(e) $x$-intercepts: $(-3, 0)$ and $(1, 0)$; $y$-intercept: $(0, -9)$.

Q3. A ball is thrown upward and outward from a height of 5 feet. The height of the ball, $f(x)$, in feet can be modeled by $f(x) = -0.1x^2 + 1.5x + 5$, where $x$ is the ball’s horizontal distance, in feet, from where it was thrown.

Background

Topic: Quadratic Applications (Projectile Motion)

This question tests your ability to interpret a quadratic function in a real-world context, specifically finding the maximum height (vertex) and the horizontal distance when the ball hits the ground ($f(x) = 0$).

Key Terms and Formulas:

• Vertex: $x = -\frac{b}{2a}$ for maximum/minimum value

• Maximum height: $f(x)$ at the vertex

• Horizontal distance when $f(x) = 0$ (solve for $x$)

Step-by-Step Guidance

1. Identify $a = -0.1$, $b = 1.5$, and $c = 5$ in the quadratic equation.

2. Find the $x$-coordinate of the vertex using $x = -\frac{b}{2a}$ to determine where the maximum height occurs.

3. Substitute this $x$ value back into $f(x)$ to find the maximum height.

4. To find how far the ball travels before hitting the ground, set $f(x) = 0$ and solve the quadratic equation $-0.1x^2 + 1.5x + 5 = 0$ for $x$ (ignore negative solutions for distance).

Try solving on your own before revealing the answer!

Final Answer:

(a) Maximum height: $f(7.5) = 10.625$ feet.

(b) The ball travels approximately $17.1$ feet horizontally before hitting the ground.

We used the vertex formula and solved the quadratic for $x$ when $f(x) = 0$.

Q4. Consider the polynomial function $f(x) = 4x^2(x-3)^2(x+2)$. Find the zeros of $f(x)$ and their multiplicities. Determine the behavior of the graph at each $x$-intercept.

Background

Topic: Polynomial Zeros and Multiplicities

This question tests your understanding of how to find zeros of a polynomial, determine their multiplicities, and describe the graph's behavior at each intercept.

Key Terms and Formulas:

• Zero: Value of $x$ where $f(x) = 0$

• Multiplicity: The exponent of the factor corresponding to the zero

• Graph behavior: Even multiplicity means the graph touches and turns; odd multiplicity means it crosses

Step-by-Step Guidance

1. Set each factor in $f(x) = 4x^2(x-3)^2(x+2)$ equal to zero to find the zeros.

2. Determine the multiplicity of each zero by looking at the exponent of each factor.

3. Describe the behavior at each $x$-intercept: if the multiplicity is even, the graph bounces; if odd, it crosses.

Try solving on your own before revealing the answer!

Final Answer:

Zeros: $x = 0$ (multiplicity 2), $x = 3$ (multiplicity 2), $x = -2$ (multiplicity 1).

At $x = 0$ and $x = 3$, the graph touches and turns; at $x = -2$, it crosses the axis.

Q5. Sketch the graph of a polynomial function $f(x)$ satisfying the following conditions: degree 6, leading coefficient $-1.7$, zeros at $x = -2$ (even), $x = 2$ (odd), $x = 5$ (odd).

Background

Topic: Graphing Polynomials with Given Zeros and Multiplicities

This question tests your ability to construct a polynomial graph based on degree, leading coefficient, and zero multiplicities.

Key Terms and Formulas:

• Degree: Highest exponent (here, 6)

• Leading coefficient: Affects end behavior

• Multiplicity: Even (touches), odd (crosses)

• End behavior: Determined by degree and sign of leading coefficient

Step-by-Step Guidance

1. Write a general form for $f(x)$ using the given zeros and their multiplicities.

2. Determine the end behavior based on the degree (even) and the negative leading coefficient.

3. At $x = -2$ (even multiplicity), the graph touches the axis; at $x = 2$ and $x = 5$ (odd), it crosses.

4. Sketch the graph, labeling zeros and showing correct end behavior.

Try solving on your own before revealing the answer!

Final Answer:

General form: $f(x) = -1.7(x+2)^{even}(x-2)^{odd}(x-5)^{odd}$ (with exponents adding to 6).

End behavior: Both ends go down; graph touches at $x = -2$, crosses at $x = 2$ and $x = 5$.