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**nycmathguy****Member**- Registered: 2021-06-02
- Posts: 53

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?

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,207

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

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**nycmathguy****Member**- Registered: 2021-06-02
- Posts: 53

Bob wrote:

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.

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