+73 Let \(f(x) = x^2 - 2 x.\)
(A) Find the slope of the secant line joining \((1, f(1))\) and \((8, f(8))\).
The slope of the secant line =
(B) Find the slope of the secant line joining \((6, f(6))\) and \((6 + h, f(6 + h))\).
The slope of the secant line =
(C) Find the slope of the tangent line at \((6, f(6))\).
The slope of the tangent line =
(D) Find the equation of the tangent line at \((6, f(6))\).
y =
f(1) = -1 f(8) = 48
find the slope of the line joining (1,-1) and (8,48) <====== do you know how to do that?
+73 I already know that the slope for part (A) is seven, and the slope for part (C) is ten, but how do I get the answer for part (B) and part (D)?
+36916 If you knew the answers to A and C.....WHY did you post them???????
For part B....you do it just the same way...with slightly different numbers ...
f(6) = 24
f(6+h) = h^2 + 12h + 36 - 12-2h = h^2 + 10h + 24
The slope between the points becomes (y1-y2) / (x1-x2)
(h^2+10h+24 -24) / (6+h - 6) = (h^2 + 10h ) / h = h+10 (check my math.....but that is how you do it.....)
+36916 For Part D ..... do you know how to do derivatives ?
f(6) = 24
find the slope at this point (6) by using the derivative
f ' (6) = 2x -2 = 10 = m
now you have a slope m = 10 and a point (6,24) to derive the equation of the line......
