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nycmathguy

Member

Registered: 2021-06-02

Posts: 53

Slope of the Tangent Line

Slope of a Tangent Line For f (x) = x^3:

(a) Find the slope of the secant line containing the points (2, 8)
and (3, 27).

Is this question asking to find m given the two points?
m = delta(y)/delta(x), where m is slope.

(b) Find the slope of the secant line containing the points (2, 8)
and (x, f (x)), x = 2.

The point (x, f(x)) = (x, x^3). Yes? Must I evaluate f(x) at x = 2? If so, what comes next?

(c) Create a table to investigate the slope of the tangent line to the
graph of f at 2 using the result from (b).

This is similar to the other thread posted about investigating the limit of a function using a table of values. Yes?

(d) On the same set of axes, graph f , the tangent line to the graph
of f at the point (2, 8), and the secant line from (a).

Since I cannot upload pictures, can someone please do part (d)?
Please, use different color lines to show the difference between the graph of f, the secant line and tangent line?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,640

Re: Slope of the Tangent Line

hi

The secant line is just the slope of AB. You can use (y2-y1)/(x2-x1)

The wording for part b is muddled.  From part c I think this is what is meant:

For the table start with x = 3 and x^3 = 27 and your result from part a.  Then let x approach 2 from above. So fill in the table with 2.5, 2.1, 2.01, 2.001 etc.  You should be able to see a limit emerging from this.

Sorry, forgot to read to the end before doing the graph so I haven't put on the tangent.  You can easily do this yourself I'm sure as part c gives the m value and you know the line goes through (2,8).

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

nycmathguy

Member

Registered: 2021-06-02

Posts: 53

Re: Slope of the Tangent Line

hi

https://i.imgur.com/ABnoIoX.gif

The secant line is just the slope of AB. You can use (y2-y1)/(x2-x1)

The wording for part b is muddled.  From part c I think this is what is meant:

For the table start with x = 3 and x^3 = 27 and your result from part a.  Then let x approach 2 from above. So fill in the table with 2.5, 2.1, 2.01, 2.001 etc.  You should be able to see a limit emerging from this.

Sorry, forgot to read to the end before doing the graph so I haven't put on the tangent.  You can easily do this yourself I'm sure as part c gives the m value and you know the line goes through (2,8).

Bob

I will further look into all 4 parts.