Question

The figure above shows the graph of the function f, defined by f(x) = $\left|2x\right|$ + 4 for all numbers x. For which of the following functions g, defined for all numbers x, does the graph of g intersect the graph of f?

• A. g(x) = x - 2
• B. g(x) = x + 3
• C. g(x) = 2x - 2
• D. g(x) = 2x + 3
• E. g(x) = 3x - 2

Hint:

You can often solve graph intersection problems visually or by comparing slopes. The vertex of f(x) is at (0, 4). To the right of the y-axis, its slope is 2. Any line with a slope greater than 2 and a y-intercept less than 4 (like g(x)=3x-2) must eventually intersect it. Any line with a slope of 2 and a lower y-intercept will be parallel and never intersect.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We are given the graph of an absolute value function, f(x) = |2x| + 4, and five linear functions. We need to determine which linear function's graph will intersect the graph of f(x). An intersection occurs if there is at least one value of x for which f(x) = g(x).
Step 2: Key Approach:
First, let's define f(x) as a piecewise function. The absolute value function |2x| can be written as:
- \(2x\) if \(x \geq 0\)
- \(-2x\) if \(x<0\)
So, the function f(x) is:
\[ f(x) = \begin{cases} 2x + 4 & \text{if } x \geq 0
-2x + 4 & \text{if } x<0 \end{cases} \] This means that for positive x, the graph of f(x) is a line with slope 2 and y-intercept 4. For negative x, it is a line with slope -2 and y-intercept 4. The vertex of this "V" shape is at (0, 4).
We can test each option g(x) by setting g(x) = f(x) for each piece of f(x) and solving for x. An intersection exists if we find a valid solution for x.
Step 3: Detailed Explanation:
Let's analyze each option.
(A) g(x) = x - 2.
- For \(x \geq 0\): \(x - 2 = 2x + 4 \implies -6 = x\). This is not valid since we assumed \(x \geq 0\).
- For \(x<0\): \(x - 2 = -2x + 4 \implies 3x = 6 \implies x = 2\). This is not valid since we assumed \(x<0\). No intersection.
(B) g(x) = x + 3.
- For \(x \geq 0\): \(x + 3 = 2x + 4 \implies -1 = x\). Not valid.
- For \(x<0\): \(x + 3 = -2x + 4 \implies 3x = 1 \implies x = 1/3\). Not valid. No intersection.
(C) g(x) = 2x - 2.
- For \(x \geq 0\): \(2x - 2 = 2x + 4 \implies -2 = 4\). This is impossible, meaning the lines are parallel. No intersection.
- For \(x<0\): \(2x - 2 = -2x + 4 \implies 4x = 6 \implies x = 1.5\). Not valid. No intersection.
(D) g(x) = 2x + 3.
- For \(x \geq 0\): \(2x + 3 = 2x + 4 \implies 3 = 4\). Impossible, lines are parallel. No intersection.
- For \(x<0\): \(2x + 3 = -2x + 4 \implies 4x = 1 \implies x = 1/4\). Not valid. No intersection.
(E) g(x) = 3x - 2.
- For \(x \geq 0\): \(3x - 2 = 2x + 4 \implies x = 6\). This is a valid solution since \(6 \geq 0\). An intersection occurs at x = 6.
- (We don't need to check the other piece since we already found an intersection, but for completeness: For \(x<0\): \(3x - 2 = -2x + 4 \implies 5x = 6 \implies x = 1.2\). Not valid.)
Step 4: Final Answer:
The graph of g(x) = 3x - 2 intersects the graph of f(x) at x = 6. Therefore, the correct answer is (E).