# a) Determine the average rate of change for the curve f(x)= -x^2 + 6x - 8 between x=2 and x=2.5 b) what is the instantaneous rate of change at x=2

beckden | High School Teacher | (Level 1) Educator

Posted on

For quadratic curves we can use

`(f(2.5)-f(1.5))/(2.5-1.5) = (0.75 - (-1.25))/1 = 2.0` Which was the answer we got above.  This only works for quadratic equations.

beckden | High School Teacher | (Level 1) Educator

Posted on

intantaneous rate of change = `lim_(h->0) (f(x+h)-f(x))/h`

`f(x+h) = -(x+h)^2 + 6(x+h) - 8 = -(x^2 + 2hx + h^2) + 6x + 6h - 8`

`f(x) = -x^2 + 6x - 8`

`f(x+h) - f(x) = -x^2 - 2hx - h^2 + 6x + 6h - 8 - (x^2 + 6x - 8) `

`= -2hx + 6h - h^2`

So `(f(x+h)-f(x))/h = (-2hx + 6h - h^2)/h = -2x + 6 - h`

`lim_(h->0) (-2x + 6 - h) = -2x + 6`

Hope that helps...

beckden | High School Teacher | (Level 1) Educator

Posted on

The average rate of change is (f(2.5)-f(2))/(2.5 - 2) = (0.75 - 0)/(0.5) = 1.5

Since the derivative of f(x) is f'(x) = -2x + 6.  The instantaneous rate of change at x=2 is f'(2) = -2(2) + 6 = 2.