College Algebra Problems With Answers
Graphs of Functions

Select points whose coordinates are easy to determine on the given graph (see graph in black below) and then transform them as follows:

1. Shift right 3 units: \( f(x - 3) \)
2. Reflect on the x-axis: \( - f(x - 3) \)
3. Shift down 2 units: \( - f(x - 3) - 2 \)
4. Connect the transformed points to sketch the transformed graph shown in red below.

A graph is symmetric with respect to the origin if for each point \((a, b)\) on the graph there exists a point \((-a, -b)\) on the same graph.

We select points \((a, b)\) on the given graph and then transform them into \((-a, -b)\) to obtain more points. When put together, the graph is symmetric with respect to the origin (see the whole graph in black and red below).

Select points on the given graph and then transform them as follows:

1. Reflect on the y-axis: \( f(-x) \)
2. Shift right one unit: \( f(- x + 1) = f(-(x - 1)) \)
3. Shift down 1 unit: \( f(- x + 1) - 1 \)
4. Connect the transformed points to sketch the transformed graph shown in red below.

We first determine points \((a, b)\) on the graph of the given function and then use the definition of the inverse to determine points \((b, a)\) on the graph of the inverse, or use the line \( y = x \) to reflect points \((a, b)\) into \((b, a)\).

• a) \( h(-2) + h(2) = -3 + 1 = -2 \)
• b) Domain: \([-5, 5]\)
• c) Range: \(\{-3\} \cup [0, 5]\)
• d) Increasing: \([-3, 2)\) and \([2, 5]\)
• e) Decreasing: \([-5, -3]\)
• f) Constant: \([-2, 2)\)

Function \( f \) has two x-intercepts: \( x = 2 \) and \( x = -2 \). \( f(2x) \) will also have x-intercepts such that \( 2x = 2 \) which gives \( x = 1 \), and \( 2x = -2 \) which gives \( x = -1 \). Hence \( f(2x) \) will have x-intercepts at \( x = 1 \) and \( x = -1 \). The y-intercept is at \( y = 2 \) since \( f(0) = 2 \) and \( f(2 \cdot 0) = f(0) = 2 \).

Since \( f^{-1}(1) = 0 \), we have \( f(0) = 1 \). Since \( f^{-1}(0) = 2 \), we have \( f(2) = 0 \).