Question

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

Answer 1

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]}