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Help with secant question , thanks

by Zephyr91
Tags: secant
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Zephyr91
#1
Jun27-07, 02:41 PM
P: 4
here is my question, *thanks for helping* i need it badly

Consider the curve y = f(x) where f(x) = (x-5)(2X+3)

A) Show that P1 = (x1,y1) = (0,-15) is on the curve and find and expression for the slope of the secant joining P1 with any other point P2 = (x2,f(x2)) , x2 cannot be 0 on the curve

b) Show that there is always a point x*, lying between x1 = 0 and x2 such that the slope of the tangent line to the curve at x* is equal to the slope of the secant joining P1 and P2



ok and i am stuck on a) how would i go about solving it, would i take the derivate of the curve then sub the points from P1 to find P2?
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Dick
#2
Jun27-07, 03:14 PM
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The secant line is just the line joining (x1,f(x1)) and (x2,f(x2)). There is no derivative involved - just a plain old slope.
daveb
#3
Jun27-07, 03:14 PM
P: 926
For A) what is the value of f(x) when you plug in x=0? Edit: OK, a little too late posting

Zephyr91
#4
Jun27-07, 03:20 PM
P: 4
Help with secant question , thanks

okay so plan old slope i would find it by doing:

m = x2 -x1 / f(x2) - f(x1)

or

slope of secant = f(0+h) - f(0) / h

m = [(o+h)-5][2(0+h)+3] - (-5)(3) / h
m = (h-5)(2h+3) + 15 / h
m = 2h^2 + 3h + 10h -15 +15 / h
m= 2h-7
?
Dick
#5
Jun27-07, 03:32 PM
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The first, except it's upside down. It's deltay/deltax.
Zephyr91
#6
Jun27-07, 03:34 PM
P: 4
do i have enough info to complete the first, i dont really know how to go about doing it...
Dick
#7
Jun27-07, 03:36 PM
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Quote Quote by Zephyr91 View Post
do i have enough info to complete the first, i dont really know how to go about doing it...
I don't know what you don't know. Are you worried that you don't know x2?
Zephyr91
#8
Jun27-07, 03:38 PM
P: 4
yeah i donty know x2
Dick
#9
Jun27-07, 03:45 PM
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Just do it anyway. What's f(x2)? Use the definition of f.


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