Help with secant question , thanks

[Zephyr91]

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?

 

[Dick]

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]

For A) what is the value of f(x) when you plug in x=0? Edit: OK, a little too late posting

 

[Zephyr91]

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]

The first, except it's upside down. It's deltay/deltax.

 

[Zephyr91]

do i have enough info to complete the first, i don't really know how to go about doing it...

 

[Dick]

do i have enough info to complete the first, i don't 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]

yeah i donty know x2

 

[Dick]

Just do it anyway. What's f(x2)? Use the definition of f.