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Questions 2
Answers 34

May 22, 2019

I guess I'll solve the second question then hahaha


To restate the important distinction in the question, \(g(x)\) is the table and \(f(x)\) is the graph.

Let's consider the validity of each option:


A: "\(f(x) \) and \(g(x)\) have the same y-intercept." For this to be true, we would want \(f(0) = g(0)\). The y-intercept of \(g(x)\) can be found in the table at \(g(0)\), which is equal to 2. Looking at the graph of \(f(x)\), we are unable to see the y-intercept at \(f(0)\), but it is definitely less than -6. Therefore, \(f(0) \neq g(0)\) and this is not true.


B: "\(f(x) \) and \(g(x)\) intersect at two points." For this to be true, we need two points from each graph that are equivalent to each other. In other words, at two \(x\) inputs, \(f(x)\) and \(g(x)\) give the same output. Let's go through each value from the table of \(g(x)\) and see if it matches \(f(x)\).\(g(-2) \neq f(-2), g(0) \neq f(0), g(2)=f(2), g(4) \neq f(2), \)and \(g(6) \neq f(6)\). We can see that \(f(x) \) and \(g(x)\) intersect at one point, but not at two. Therefore, this is not true.


C: "\(f(x)\) is greater than \(g(x)\) for all values of \(x\)." We can quickly find a counterexample for this. At \(x=4\)\(g(4) = 2\) and \(f(4) = 0\). Therefore, \(f(4) \ngtr g(4)\) and this is not true.


D: "\(f(x) \) and \(g(x)\) have a common x-intercept." We learned from looking at B that \(g(2) = f(2)\), wherein both equaled a y-value of \(0\). Therefore, this is true.


Our answer to this question is then the fourth option, or as I have labeled it, D.

May 17, 2019

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May 14, 2019