Question

write the equation of the line tangent to the graph of f(x) at x=-1. f(x)=6x^2+9x

Answer 1

the line tangent to f(x) at x = -1 has the same slope as the graph of f(x) at that point x = -1.

You can find the slope of f(x) by finding its derivative... then find the slope specifically at x = -1 by substituting x = -1 into the expression of the derivative, f '(x)

Answer 2

i tried that but i got -3 and the answer to the question is y=-3x-6 and i dont understand how to get that

Answer 3

tangent slope is f'(x) = 12x + 9

f ' (x) at x = -1 is 12(-1) + 9 = -3

The value of f(x) at x = -1 is f(-1) = 6(-1)^2 + 9(-1) = -3, so the ordered pair is (-1,-3).

Now you have point and slope... enough to write the tangent line equation.

y - y1 = m (x - x1) equation of line in point slope format
y - (-3) = -3(x - (-1)
y + 3 = -3x -3
y = -3x - 6

Double check:
slope of tangent line = -3 (yes)
Point on f(x) at f(-1) is (-1,-3) which also satisfies y = -3x - 6 (yes)

Answer 4

when you got "-3" that was just the slope of the tangent line, not the actual equation of the tangent line...

Answer 5

you started fine, but you didn't finish the problem all the way :) Read the explanation above and let me know if you have questions about it.

Answer 6

okay thank you so much i get it now! (:

Answer 7

great!