Question

f(x) = -4^3 - 15x^2 + 18x - 5
Write equation of the tangent line at x = -1

Answer 1

@jim_thompson5910 Can you help? c:

Answer 2

First derive with respect to x

f(x) = -4x^3 - 15x^2 + 18x - 5

f ' (x) = -3*4x^2 - 2*15x^1 + 18

f ' (x) = -12x^2 - 30x + 18

Answer 3

plug in x = -1 into f ' (x) = -12x^2 - 30x + 18 to find the slope of the tangent line at x = -1

this will give you the value of m

Answer 4

tell me what you get for m

Answer 5

36

Answer 6

the slope of the tangent line at x = -1 is m = 36, good

Answer 7

now tell me what you get when you plug x = -1 into the original function f(x)

Answer 8

-34

Answer 9

this means the point (-1, -34) lies on the graph of the function f(x)

Answer 10

you now have a slope and a point...all the necessary ingredients to find the equation of a line

Answer 11

y = 36x + 2

Answer 12

that's correct

Answer 13

nice work

Answer 14

Awesome! :O Well, ok. How about this,
Find the x value(s) where the slope of the tangent line is 0

Answer 15

plug in f ' (x) = 0 and solve for x

f ' (x) = -12x^2 - 30x + 18

0 = -12x^2 - 30x + 18

-12x^2 - 30x + 18 = 0

...

x = ?? or x = ??

Answer 16

OHHHHHHHHHHHHHHHHHHHHHHH -3 and -0.5

Answer 17

one of those is wrong

Answer 18

\[\frac{ 5\pm \sqrt{5^2 - 4(-2)(3)} }{ 2(-2) }\]

Answer 19

Is there anything wrong with the quadratic formula?

Answer 20

you should have (5+7)/(-4) and (5-7)/(-4)

Answer 21

Oh I got it, the 0.5 isn't negative

Answer 22

good

Answer 23

So if they tell me to find the Max/Min points I just use the x values and plug it back in to the original formula to find y, then I'd have the points.

Answer 24

mmhmm

Answer 25

you have to use the first or second derivative test though

Answer 26

So max would be (-0.25,0.5) min would be (-3,-86)

Answer 27

What is that o.o

Answer 28

The first derivative tells you where the function is increasing and decreasing using the critical points, or the "zeroes" of a function

The second derivative tells you where the function is concave up or down,and the inflection points where the original function changes concavity.

Answer 29

I'm sorry, but can you dumb it down a bit? Lol

Answer 30

lets see...