- #1
Zephyr91
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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?
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?