+0

# Let f(x) = x^2 - 2x.

+1
175
5

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 =

Feb 25, 2022

#1
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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?

Feb 25, 2022
#2
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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)?

GAMEMASTERX40  Feb 25, 2022
#3
+2

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.....)

Feb 25, 2022
edited by ElectricPavlov  Feb 25, 2022
edited by ElectricPavlov  Feb 25, 2022
#4
+2

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......

Feb 25, 2022
#5
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Gamemaster,

You need to start talking about the questions that you post.

I mean you need to discuss what you already think you know and what thoughts you have on the rest.

Just dumping questions is simply NOT cool.

I think you are capable of better. Feb 26, 2022