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# I need help, but it's not urgent haha :)

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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
+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" )

Nov 15, 2019
#2
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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
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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   Nov 15, 2019
#4
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Thanks Phil! I get it now lol :))

Nov 18, 2019