College Algebra Review: Step-by-Step Guidance for Key Topics

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 value of $a$ in the quadratic function. Recall that if $a > 0$, the parabola opens upward; if $a < 0$, it opens downward.

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

3. Write the equation for the axis of symmetry using the $x$-coordinate of the vertex.

4. To find the $x$-intercepts, set $f(x) = 0$ and solve the resulting quadratic equation for $x$.

5. 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: $(0, 0)$.

(e) Use these points and the axis of symmetry to sketch the parabola, labeling the vertex, intercepts, and axis of symmetry.

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

Background

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

This question tests your ability to analyze and graph a quadratic function, including converting to vertex form, finding intercepts, and determining domain and range.

Key Terms and Formulas:

• Vertex form: $f(x) = a(x-h)^2 + k$

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

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

• Intercepts: Solve $f(x) = 0$ for $x$-intercepts; $f(0)$ for $y$-intercept

Step-by-Step Guidance

1. Determine if the parabola opens upward or downward by checking the sign of $a$.

2. Find the vertex using $x = -\frac{b}{2a}$ and substitute back to get the $y$-coordinate.

3. Rewrite $f(x)$ 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 $f(x) = 0$ 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 model in a real-world context, specifically finding the maximum value and the zeros of the function.

Key Terms and Formulas:

• Maximum height: The vertex of the parabola (since $a < 0$)

• Horizontal distance before hitting the ground: Solve $f(x) = 0$ for $x$

Step-by-Step Guidance

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

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

3. To find how far the ball travels before hitting the ground, set $f(x) = 0$ and solve the quadratic equation for $x$ (ignore negative solutions).

Try solving on your own before revealing the answer!

Final Answer:

(a) Maximum height: 6.125 feet (at $x = 7.5$ feet).

(b) The ball travels approximately 16.2 feet horizontally before hitting the ground.

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 Functions (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 on the factor corresponding to the zero

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

Step-by-Step Guidance

1. Set each factor equal to zero to find the zeros: $x^2 = 0$, $(x-3)^2 = 0$, $x+2 = 0$.

2. Determine the multiplicity of each zero by looking at the exponent on 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 Polynomial Functions (End Behavior, Zeros, Multiplicities)

This question tests your ability to use information about degree, leading coefficient, and zeros to sketch a polynomial graph.

Key Terms and Formulas:

• Degree: Determines end behavior

• Leading coefficient: Affects whether ends go up or down

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

Step-by-Step Guidance

1. Recall that a degree 6 polynomial with a negative leading coefficient will have both ends pointing downward.

2. At $x = -2$ (even multiplicity), the graph touches the axis and turns.

3. At $x = 2$ and $x = 5$ (odd multiplicities), the graph crosses the axis at these points.

4. Sketch the graph, making sure to reflect the correct end behavior and behavior at each zero.

Try sketching the graph on your own before checking the answer!

Final Answer:

Graph has both ends down, touches at $x = -2$, crosses at $x = 2$ and $x = 5$.