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

graph of quadratic function f of t having x intercepts at 6, 0 and 18, 0 and vertex at 12, negative 36 and passes through 3, 49.5 and 9, negative 27.25

What is the closest approximate average rate of change for Edna's profits from the 3rd month to the 9th month?

A. −69.75 dollars per month

B. −11.63 dollars per month

C. Six dollars per month

D. Nine dollars per month

(i think it's C but i'm not sure)

Respuesta :

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

[tex](x-6)(x-18)=x^{2}-6x-18x+108[/tex]
[tex]=x^{2}-24x+108[/tex]

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

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

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.
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markguevara03

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

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