Answer:
Function A has the greater average rate of change over the interval [1, 2]
Explanation:
Given the below functions and interval [1, 2};
[tex]\begin{gathered} \text{Function A: }g(x)=x^2+4x-8 \\ \text{Function B: }(x)=x^2-3x+6 \end{gathered}[/tex]
Note that the below formula can be used to determine the average rate of change over an interval [a, b};
[tex]\frac{f(b)-f(a)}{b-a}[/tex]
Average rate of change of Function A:
Given the interval [1, 2] where a = 1 and b = 2, let's go ahead and determine g(a), g(b), and the average rate of change of Function A as seen below;
[tex]\begin{gathered} g(1)=(1)^2+4(1)-8=1+4-8=-3 \\ g(2)=(2)^2+4(2)-8=4+8-8=4 \\ \text{Average rate of change }=\frac{4-(-3)}{2-1}=\frac{4+3}{1}=\frac{7}{1}=7 \end{gathered}[/tex]
So the average rate of change of Function A is 7
Average rate of change of Function B:
Given the interval [1, 2] where a = 1 and b = 2, let's go ahead and determine h(a), h(b), and the average rate of change of Function B as seen below ;
[tex]\begin{gathered} h(1)=(1)^2-3(1)+6=1-3+6=4 \\ h(2)=(2)^2-3(2)+6=4-6+6=4 \\ \text{Averate rate of change }=\frac{4-4}{2-1}=\frac{0}{1}=0 \end{gathered}[/tex]
So the average rate of change of Function B is 0
If we compare the average rate of change of the two functions, we can see that Function A has the greater average rate of change over the interval [1, 2]
