Question

The graph below shows the value of Edna's profits f(t), in dollars, after t months:

Answer 1

Since we know that the graph of our quadratic function has x-intercepts at (6,0) and (18,0), $x=6$ and $x=18$ are the zeroes of our quadratic. To find our quadratic we are going to factor each zero backwards and multiply them:
$x=6$
$x-6=0$
$x=18$
$x-18=0$

$(x-6)(x-18)=x^{2}-6x-18x+108$
$=x^{2}-24x+108$

Now that we have our quadratic function, we are going to use the average formula: $m= \frac{f(9)-f(3)}{9-3}$
$where$
$f(9)$ is the function evaluated at  the 9th month
$f(3)$ is the function evaluated at the 3rd month 

$m= \frac{f(9)-f(3)}{9-3}$
$m= \frac{[9^{2}-24(9)+108]-[3^{2}-24(3)+108]}{9-3}$
$m= \frac{-27-45}{6}$
$m= \frac{-72}{6}$
$m=-12$

We can conclude that the closest approximate average rate of change for Edna's profits from the 3rd month to the 9th month is: B. −11.63 dollars per month.

Answer 2

I got -11, so I probably did have a bit of math errors, so I'm pretty sure then that is is -11.63