College Algebra Test 3 Review – Step-by-Step Guidance

Q1. Use the vertex and intercepts to sketch the graph of the quadratic function: $f(x) = 4(x - 6)^2 + 4$

Background

Topic: Quadratic Functions and Graphing

This question tests your ability to analyze and graph a quadratic function using its vertex form, and to find its intercepts for sketching.

Key Terms and Formulas:

• Vertex form of a quadratic: $f(x) = a(x - h)^2 + k$

• Vertex: $(h, k)$

• Axis of symmetry: $x = h$

• Y-intercept: $f(0)$

• X-intercepts: Solve $f(x) = 0$

Step-by-Step Guidance

1. Identify the vertex from the equation. The function is in vertex form, so $h = 6$ and $k = 4$. The vertex is $(6, 4)$.

2. Determine the direction the parabola opens. Since $a = 4$ (positive), the parabola opens upward.

3. Find the y-intercept by evaluating $f(0)$: $f(0) = 4(0 - 6)^2 + 4$.

4. Find the x-intercepts by setting $f(x) = 0$ and solving $4(x - 6)^2 + 4 = 0$.

Try solving on your own before revealing the answer!

Final Answer:

The vertex is at $(6, 4)$. The y-intercept is at $(0, 148)$. The x-intercepts are at $x = 6 \pm 0.5$.

The graph is a parabola opening upward, centered at $(6, 4)$, and crosses the y-axis at $(0, 148)$.

Q2. Use the vertex and intercepts to sketch the graph of the quadratic function: $f(x) = (x - 3)^2 - 1$

Background

Topic: Quadratic Functions and Graphing

This question tests your ability to analyze and graph a quadratic function in vertex form, and to find its intercepts for sketching.

Key Terms and Formulas:

• Vertex form: $f(x) = a(x - h)^2 + k$

• Vertex: $(h, k)$

• Y-intercept: $f(0)$

• X-intercepts: Solve $f(x) = 0$

Step-by-Step Guidance

1. Identify the vertex: $h = 3$, $k = -1$, so the vertex is $(3, -1)$.

2. Determine the direction the parabola opens. $a = 1$ (positive), so it opens upward.

3. Find the y-intercept: $f(0) = (0 - 3)^2 - 1$.

4. Find the x-intercepts: Set $(x - 3)^2 - 1 = 0$ and solve for $x$.

Try solving on your own before revealing the answer!

Final Answer:

The vertex is at $(3, -1)$. The y-intercept is at $(0, 8)$. The x-intercepts are at $x = 3 \pm 1$.

The graph is a parabola opening upward, centered at $(3, -1)$, and crosses the y-axis at $(0, 8)$.

Q3. Use the vertex and intercepts to sketch the graph of the quadratic function: $f(x) = -x^2 - 2x + 8$

Background

Topic: Quadratic Functions and Graphing

This question tests your ability to analyze and graph a quadratic function in standard form, and to find its vertex and intercepts for sketching.

Key Terms and Formulas:

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

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

• Y-intercept: $f(0)$

• X-intercepts: Solve $f(x) = 0$

Step-by-Step Guidance

1. Find the vertex using $x = -\frac{b}{2a}$: $a = -1$, $b = -2$.

2. Calculate $y$ value of the vertex by plugging $x$ into $f(x)$.

3. Determine the direction the parabola opens. $a = -1$ (negative), so it opens downward.

4. Find the y-intercept: $f(0) = 8$.

5. Find the x-intercepts by solving $-x^2 - 2x + 8 = 0$.

Try solving on your own before revealing the answer!

Final Answer:

The vertex is at $(-1, 9)$. The y-intercept is at $(0, 8)$. The x-intercepts are at $x = -4$ and $x = 2$.

The graph is a parabola opening downward, centered at $(-1, 9)$, and crosses the y-axis at $(0, 8)$.