Precalculus 12 grade

Question

F(x)= 5x^2-11x-9 . find average rate from a=2 to a=2+h

Answer 1

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

is an upward opening parabola

f(x) = 5(x^2 -11x/5 + 121/100) - 9 + 121/100

= 5(x-11/10)^2 -7.79 the vertex is (1.1, -7.79)

for any x value greater than 1.1, the average rate of increase will be positive and it increases more as x increases

at x =2, f'(2) = 10(2)-11 = 9 = the instantaneous rate of change at the point (2,-11)

at any x=a >2, the rate of change >9

to jump to the solution the average rate of change = 5h+9. if h=0 the that's the same as the instantaneous rate of change when x=a = 2, = 5(0)+9 = 9

the average rate of change from a=2 to a=2+h is (f(2+h)-f(2))/h = 5h+9

average rate of change is the slope of the secant line from (2, f(2)) to (2+h, f(2+h))

as h approaches zero, the secant line slope approaches the tangent line slope = 9

Answer 2

Average rate is essentially the slope formula (y₂ - y₁)/(x₂ - x₁) except we fancy it up by putting it in function format:

(f(x₂) - f(x₁))/(x₂ - x₁) or, in this case: (f(a + h) - f(a))/((a + h) - a)

f(a + h) = 5(a + h)² - 11(a + h) - 9 = 5(a² + 2ah + h²) - 11a - 11h - 9 = 5a² + 10ah + 5h² - 11a - 11h - 9

f(a) = 5a² - 11a - 9

So f(a + h) - f(a) = (5a² + 10ah + 5h² - 11a - 11h - 9) - (5a² - 11a - 9) = 10ah + 5h² - 11h (some things cancel out)

And the denominator simplifies to "h"

So, the average rate becomes:

(10ah + 5h² - 11h)/h = h(10a + 5h - 11)/h = 10a + 5h - 11

Since a = 2, 10(2) + 5h - 11 = 9 + 5h or 5h + 9