SOLUTION: https://gyazo.com/476623f75a5bf00a7c1a4802c602403e

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Question 1178952: https://gyazo.com/476623f75a5bf00a7c1a4802c602403e
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

y=%283x%5E2-4x%29%2F%282x%2B1%29 tangent at+x=-1

first find a slope and tangent point

y'%28x%29+=+%286x%5E2+%2B+6+x+-+4%29%2F%282+x+%2B+1%29%5E2
substitute x=-1
y'%28-1%29+=%286%28-1%29%5E2+%2B+6+%28-1%29+-+4%29%2F%282+%28-1%29+%2B+1%29%5E2
y'%28-1%29+=+-4->Note this only tells us the slope of the line, not the equation for the line itself.

since the line is tangent to the function at +x=-1, by finding the y-coordinate we would possess a set of coordinates and a slope, allowing us to use point-slope form

y=%283%2A%28-1%29%5E2-4%2A%28-1%29%29%2F%282%2A%28-1%29%2B1%29+=-7
tangent point is (-1,-7) and slope ism=+-4

use point-slope form

y-y%5B1%5D=m%28x-x%5B1%5D%29
y-%28-7%29=-4%28x-%28-1%29%29
y%2B7=-4%28x%2B1%29
y=-4x-4-7
y=-4x-11