## Help with secant question , thanks

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

## 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
?
 Recognitions: Homework Help Science Advisor The first, except it's upside down. It's deltay/deltax.
 do i have enough info to complete the first, i dont really know how to go about doing it...

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