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March 6, 2015

March 6, 2015

Posted by **Yoona** on Monday, November 7, 2011 at 12:09am.

a. Find the zeros of f

b. Write an equation of the line tangent to the graph of f at x = -1

c. Find the x coordinate of the point where the tangent line is parallel to the secant line on the interval [1, 3].

- calculus -
**Damon**, Monday, November 7, 2011 at 1:48amWell, first off, if x=1 we have a zero.

So divide by (x-1) and get

x^2+x-6

factor that

(x-2)(x+3)

so the zeros are at x = 1, 2, -3

At x - -1

y = -1 + 7 + 6 = 12

y' = 3 x^2 - 7 which is -4 at x = -1

so this line goes through (-1,12) and has slope m = -4

You can do the rest

at x = 1, y = 0

at x = 3, y = 27-21+6 = 12

line slope m = 12/2 = 6

where is slope of tangent line = 6?

3x^2 -7 = 6

3 x^2 = 13

x^2 = 13/3

x = sqrt (13/3)

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