College Algebra Exit Exam Review: Step-by-Step Guidance

Q14b. Graph each equation and give the domain and range: $g(x) = -|x|$

Background

Topic: Absolute Value Functions

This question tests your understanding of how to graph absolute value functions, determine their domain and range, and interpret their features. The negative sign in front of the absolute value changes the direction in which the graph opens.

Key Terms and Formulas:

• Absolute value function: $f(x) = |x|$

• Domain: The set of all possible input values ($x$) for the function.

• Range: The set of all possible output values ($y$) for the function.

• For $g(x) = -|x|$, the graph is a 'V' shape opening downward.

Step-by-Step Guidance

1. Recall the basic graph of $f(x) = |x|$, which is a 'V' shape opening upward with its vertex at the origin $(0,0)$.

2. Apply the negative sign: $g(x) = -|x|$ reflects the graph of $|x|$ over the $x$-axis, so the 'V' now opens downward.

3. Identify the domain: Since absolute value functions are defined for all real numbers, $g(x)$ is also defined for all real $x$.

4. Consider the range: Because the graph opens downward, the maximum value is at the vertex, and all other values are less than or equal to this maximum.

Try solving on your own before revealing the answer!

Final Answer:

Domain: $(-\infty, \infty)$

Range: $(-\infty, 0]$

The graph is a downward opening 'V' with vertex at $(0,0)$, and the function is defined for all real $x$ with outputs less than or equal to zero.

Q15a. Graph each parabola. Identify the vertex and y-intercept: $f(x) = -(x+1)^2 + 4$

Background

Topic: Quadratic Functions and Parabolas

This question tests your ability to graph a quadratic function in vertex form, identify the vertex, and find the y-intercept. The negative sign indicates the parabola opens downward.

Key Terms and Formulas:

• Vertex form: $f(x) = a(x-h)^2 + k$

• Vertex: $(h, k)$

• y-intercept: The value of $f(x)$ when $x = 0$

• If $a < 0$, the parabola opens downward.

Step-by-Step Guidance

1. Identify the vertex from the equation: $f(x) = -(x+1)^2 + 4$ is in vertex form, so $h = -1$, $k = 4$.

2. Determine the direction: Since $a = -1$, the parabola opens downward.

3. Find the y-intercept by plugging $x = 0$ into the equation: $f(0) = -(0+1)^2 + 4$.

4. Sketch the graph using the vertex and y-intercept, and note the symmetry about $x = -1$.

Try solving on your own before revealing the answer!

Final Answer:

Vertex: $(-1, 4)$

y-intercept: $(0, 3)$

The parabola opens downward, with its highest point at $(-1, 4)$ and crosses the y-axis at $(0, 3)$.

Q15b. Graph each parabola. Identify the vertex and y-intercept: $f(x) = -6x^2 + 4x$

Background

Topic: Quadratic Functions and Parabolas

This question tests your ability to graph a quadratic function in standard form, find the vertex using the vertex formula, and determine the y-intercept.

Key Terms and Formulas:

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

• Vertex formula: $x = -\frac{b}{2a}$

• y-intercept: The value of $f(x)$ when $x = 0$

• If $a < 0$, the parabola opens downward.

Step-by-Step Guidance

1. Identify coefficients: $a = -6$, $b = 4$, $c = 0$.

2. Find the x-coordinate of the vertex using $x = -\frac{b}{2a}$.

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

4. Find the y-intercept by evaluating $f(0)$.

Try solving on your own before revealing the answer!

Final Answer:

Vertex: $(\frac{1}{3}, \frac{4}{3})$

y-intercept: $(0, 0)$

The parabola opens downward, with its vertex at $(\frac{1}{3}, \frac{4}{3})$ and crosses the y-axis at $(0, 0)$.