# Help with secant question , thanks

by Zephyr91
Tags: secant
 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?
 Sci Advisor HW Helper Thanks P: 25,228 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.
 P: 925 For A) what is the value of f(x) when you plug in x=0? Edit: OK, a little too late posting
 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 ?
 Sci Advisor HW Helper Thanks P: 25,228 The first, except it's upside down. It's deltay/deltax.
 P: 4 do i have enough info to complete the first, i dont really know how to go about doing it...