Precalculus 12 grade.

Question

F(X)=5x^2-11x-9, find the average rate of change from a=2 to a=2+h

Answer 1

I see tutor William W. answered and did so excellently in my opinion.

but, alternative way is f(2)=-11

f(2+h)=20+20h+5h²-22-11h-9=-11+9h+5h²

f(2+h)-f(2)=-5h²+9h

that is change in f(x) for x=2 to x=2+h

change in x is h

(change in f(x))/(change in x)=5h+9

for x=2 to x=2+h

hey, what happens if h=1. that is, average rate of change (slope) of f(x) from 2 to 3

it is no longer dy/dx because h does not approach zero, let's call value m

m=5h+9=14, meaning the line joining (2,f(2)) and (3,f(3)) has slope m=14

f(3)=3, f(2)=-11,,,(f(3)-f(2))/(3-2)=14,,,,formula works

try h=2.,,,,m=5h+9=19.

f(4)=27, f(4)-f(2)=38,,,,38/(4-2)=19,,,,formula works

try h=-1 m=5h+9=4

f(1)=-15,,,f(1)-f(2)=-4,,,,-4/(1-2)=4,,,,formula works

Answer 2

f(x) = 5x^2 -11x -9

average rate of change from a=2 to a=2+h?

calculate [f(x+h) -f(x)]/h

[f(2+h) -f(2)]/h

= 5(2+h)^2 -11(2+h) -9 - [5(2)^2 -11(2) -9]/h

= [20+ 20h +5h^2 -22 -11h -20 +22]/h

= (5h^2 + 9h/)h=9

= 5h +9= slope of secant line from x=2 to x=2+h

as h approaches zero, the slope approaches 9

= average rate of change between points (2,f(x)) and (2+h, f(2+h))

f(x)=5x^2 -11x -9

f'(x) = 10x -11= slope

f'(2) =10(2) - 11 = 20-11 = 9 = slope of tangent line when x=2

= instantaneous rate of change when x=2

f(2+h) = 10(2+h)-11 = 20+20h-11 = 20h+9= slope of tangent line when x=2+h

f(x) is an upward opening parabola, with increasing slope to the right of the vertex

5x^2 -11x -9

= 5(x^2 -11x/5 + (11/10)^2) -9 - 5(11/10)^2

= 5(x- 11/10)^2 -900/100 - 5(121)/100

= 5(x-1.1)^2 - 1505/100

=5(x-1.1)^2 -15.05

vertex is (1.1, -15.05)

for x>1.1, slope is increasing