Request if you can tell me where i’m going wrong or direct me to a gregmat video that can help.
You can solve this algebraically by equating f(x) and g(x) and finding the x values that solve that problem.
For example, f(x) = g(x) → |2x| + 4 = x +3 → |2x| = x - 1.
Then, we get two separate equations to solve: (a) 2x = x - 1 & (b) 2x = -(x-1)
For (a), 2x = x - 1 → x = -1. However, for absolute value equations, we always need to plug our answer back into the original equation to check if it is possible.
We see if we plug in x = -1 into |2x| + 4 = x + 3, then the left hand side is |2(-1)| + 4 = |-2| + 4 = 2 + 4 = 6 while the right hand side is -1 + 3 = 2. So, x = - 1 does not work. We could continue the process, but you may see that this will be a long brute force calculation.
Instead, graphically, we can see that choice (A) will lie underneath f(x) and will never intersect because its y-intercept is smaller and for positive x values, the slope 1 will be less than f(x)'s slope of 2, so g(x) will never reach f(x). Similar reasoning works for choice (B).
For choices (C) & (D), we can see that the slope is still a problem because even though the slopes are equal for the positive x values, the y-intercept of g(x) is smaller than the y-intercept of 4 for f(x). Thus, g(x) will still never reach f(x). These graphs will be parallel for positive x values.
In both cases before, we disregarded an intersection when x < 0 because the graph of g(x) will clearly be below f(x). Feel free to plot out these graphs and see.
Lastly, choice (E) is correct because the graphs will intersect for some positive x-value. Even though, g(x) starts at a lower y-intercept, it will eventually catch up and overtake f(x) because g(x) has a slope of 3, which is greater than the slope of 2 for f(x) when x > 0.
I would review graph transformation rules and absolute value equations on the GregMat website. Feel free to look at the 1 month plan Flashcards. Those can be helpful.
I hope this helps!
Thank you!
