find slope of secant line

Question

For the function f(x)=4x^2, make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at x=3

interval [2,3] 

find slope of secant line 

Answer 1

It helps if you draw the parabola

and the tangent and a couple secant lines

just a rough sketch helps a lot

f(x)=4x^2

slope of tangent line = derivative = f'(x) = 8x

f'(3) = 8(3) = 24 = slope of tangent line at the point (2,16)

slope of secant line on the interval x=2 to x=3 is "rise over run" = change in y over change in x =

= (f(3)-f(2))/(3-2) = 36-16=20

slope of secant line on the interval x=2.5 to x=3

= (f(3)-f(2.5))/(3-2.5) = (36-25)/.5 = 11/.5 = 22

secant line slope goes from 20 to 22 and likely to 24 as the interval approaches zero

limit of the secant line slope = the tangent line slope = 24 at x=3

if you calculated "rise over run" on "interval" x =3 to x=3 you get an undefined 0/0

= (f(3)-f(3))/(3-3) = (36-36)/0 = 0/0