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avatar+279 

Given the function f(x) = 4x^2 - 10x + 9, What is the average rate of change for f ove rthe interval 2.6 le x le 5.2?

 

(le means less than or equal too btw :))

 

Thank you so, so much if you can take any time to help me, if not that's ok too though

 Nov 15, 2019
 #1
avatar+8829 
+2

 

The average rate of change of  f  over the interval  2.6 ≤ x ≤ 5.2  is the slope of the blue dotted line on the above graph.

 

The blue dotted line passes through the purple point,  (2.6, f(2.6))   , and the green point,  (5.2, f(5.2))

 

Do you know how to find the slope of the blue dotted line? (Remember "rise over run" smiley)

 Nov 15, 2019
 #2
avatar+279 
0

Uhhh I think I did something wrong :PP I got (15-5)\(3-2) = 10 as the slope but I don't think that's right.. :((

 Nov 15, 2019
 #3
avatar+105411 
+2

f(2.6)  =  4(2.6)^2  - 10(2.6) + 9  = 10.04

f(5.2)  = 4(5.2)^2 - 10(5.2) + 9  =65.16

 

So    we have the points   ( 2.6, 10.04)  and  ( 5.2, 65.16)

 

The slope is

 

[ 65.16 - 10.04 ]  / [ 5.2 - 2.6 ]   = 21.2  = avg rate of change

 

 

cool cool cool

 Nov 15, 2019
 #4
avatar+279 
0

Thanks Phil! I get it now lol :))

 Nov 18, 2019

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