Question

HELPPP! determine the y-intercept of a line with a slope of -2 that is tangent to the curve y=1/3x^2 - 4x -3

Answer 1

first find the derivative y':
y'=(2/3)x - 4

this gives the slope of the curve at any point x. since we want the line to have slope -2, we need to solve:

-2 = (2/3)x - 4
2 = (2/3)x
3 = x

this is the x coordinate on the graph where the slope is -2.
we have slope, now we need a point on that line. what better point than the point of tagency....

y = (1/3)(3^2) - 4(3) - 3
= 3 -12 - 3 = -12

slope = -2 through (3, -12), the line in point slope form is: y+12=-2(x-3)

i trust you can do the rest...?