Slope of Secant Homework: y=x^2+x, P(1,2) & Q(2,6)

[physphys]

Homework Statement

1. For the curve y = x^2+ x, and the point P (1,2)
a) determine the slope of the secant mPQ if Q(2, 6)
b) what is the slope of the secant if Q is at the following values of x
1.5 1.1 1.01 1.001
c) Estimate the value of the tangent to the curve at P. Reason?

Homework Equations

Y'= f(x+h)-f(x)
------------
h

The Attempt at a Solution

If i plug in the numbers into the y' equation i would get

y' = (2+h)^2 + 2 + h - 6
---------------------
h
which is h^2 + 5
and then h-> 0 so h = 5

I'm not sure if that is right, because isn't slope Δy/Δx ? and doing that would give me 4.
and when I use the slope function 2x+1, at point 2, 2(2)+1 = 5.
So I'm very confused between the difference in finding out the tangent line or the secant line. If you can just show me how to find the secant line in a) i can figure out b) on my own. Thanks!

 

[Mark44]

Homework Statement

1. For the curve y = x^2+ x, and the point P (1,2)
a) determine the slope of the secant mPQ if Q(2, 6)
b) what is the slope of the secant if Q is at the following values of x
1.5 1.1 1.01 1.001
c) Estimate the value of the tangent to the curve at P. Reason?

Homework Equations

Y'= f(x+h)-f(x)
------------
h

The Attempt at a Solution

If i plug in the numbers into the y' equation i would get

y' = (2+h)^2 + 2 + h - 6
---------------------
h
which is h^2 + 5
and then h-> 0 so h = 5

I'm not sure if that is right, because isn't slope Δy/Δx ? and doing that would give me 4.
and when I use the slope function 2x+1, at point 2, 2(2)+1 = 5.
So I'm very confused between the difference in finding out the tangent line or the secant line. If you can just show me how to find the secant line in a) i can figure out b) on my own. Thanks!

Your answer of 4 for part a is correct.

If you draw a sketch of the curve y = x² + x, and draw the tangent line at (1, 2) and the secant line between (1, 2) and (2, 6), that should help your understanding.

 

[sleepycoffee]

You could think of the secant as the average slope, which you get from rise/run=4/1=4, which you did correctly.

With lim h->0 (f(x+h)-f(x)) / h, you are calculating the slope of the tangent line at f(x). So, when you calculated y', you found found the tangent at point P.

For b, you would do the same thing as in a. secant=(f(1.5)-f(1))/0.5, (f(1.1)-f(1))/0.1, and so on. What you should find is that as the difference gets smaller (0.5->0.1->0.01 etc), the value of the secant approaches the value of the tangent (that is, it should get closer and closer to 5).

Hope that helps.

 

[Whovian]

I'm not sure if that is right, because isn't slope Δy/Δx ? and doing that would give me 4.
and when I use the slope function 2x+1, at point 2, 2(2)+1 = 5.

(Pulled these images off Wikipedia.)

Notice how the secant line is the line connecting two points. In this case, the secant line between (1,2) and (2,6) would have slope Δy/Δx.

The slope of the tangent line at a point P is the slope of the secant line PQ as Q approaches P (while being on the curve.) In this sense, the tangent line's slope is independent of the value of (2,f(2)).

 

[physphys]

Okay understood! thanks to all for your help!