Well, 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)
5. Let f be the function given by f(x) = x3- 7x + 6.
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].
1 answer
1 answer