How to find the indicated quantities for f(x) = 3x^2?

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1 Answer
Feb 23, 2017

A) Slope = #6+3h#
B) Slope = 6
C) # y = 6x-3 #

Explanation:

We have # f(x) = 3x^2 #

A. Slope of Secant Line
The slop of the secant line is given by:

# (Delta y)/(Delta x) = {f(1+h)-f(1)}/{(1+h)-(1)}#
# \ \ \ \ \ \ \ = {3(1+h)^2-3}/{h}#
# \ \ \ \ \ \ \ = {3(1+2h+h^2)-3}/{h}#
# \ \ \ \ \ \ \ = {3+6h+3h^2-3}/{h}#
# \ \ \ \ \ \ \ = {6h+3h^2}/{h}#

# \ \ \ \ \ \ \ = 6+3h#

B. Slope of the graph (tangent) at #(1.f(1))#
If wd take the limit of the slope at the secant line (A) then by the definition of the derivative then in the limit as #h rarr 0# this will become the slope of the tangent at #x=1#, So

# lim_(h rarr 0) 6+3h = 6 #

C. Equation of tangent
Th slope of the tangent is #6#, and it passes through #(1,3)# so using the point/slope equation of a straight line #y-y_1=m(x-x_1)# we get:

# \ \ \ \ \ y-3 = 6(x-1) #
# :. y-3 = 6x-6 #
# :. \ \ \ \ \ \ \ y = 6x-3 #

Which we can confirm via a graph:
graph{ (y-3x^2)(y-6x+3)=0 [-5, 5, -2, 10]}