Question

function is: 9x^3+10x^2+5
find the equation of the tangent line at x=0, please show steps

Answer 1

can you find the 1st derivative...?

Answer 2

Yup

Answer 3

ok... so what did you get...?

Answer 4

the derivative stuff isn't the problem. I'm only slightly confused on a certain part

Answer 5

the 1st derivative gives the equation for the slope of the tangent...

Answer 6

Yes, yes I understand that

Answer 7

ok...

so you have

\[\frac{dy}{dx} = 27x^2 + 20x\]

Answer 8

yes

Answer 9

ok... so find the slope of the tangent at x = 0

substitute x = 0 into the 1st derivative...

so m = 0

you are dealing with a horizontal line

does that make sense..?

Answer 10

Yes, I've gotten all that

Answer 11

I just wasn't sure if that was correct. It just didn't look right

Answer 12

ok.. you need to y value for the point where x = 0

so substitute x = 0 into the original equation

y = 9*0^3 + 10*0^2 + 5

so y = 5

the point you are working with is

(0, 5)

does that make sense..?

Answer 13

but since you confirmed my answer, I guess there's not a real problem

Answer 14

Yep, I have all that :)

Answer 15

well the slope intercept form is

y = mx + b

in this question y = b

substitute x = 0 and y = 5

gives b = 5

so the line is y = 5

Answer 16

Ok good good yes yes I have thees :D

Answer 17

w00t

Answer 18

well done...

Answer 19

wouldn't it be -5 (unless I'm crazy)

Answer 20

never mind

Answer 21

I goofed

Answer 22

I think its y = 5

are you happy with the point being (0, 5)...?

and m = 0

so y = 0x + b

or 5 = 0*0 + b

b = 5

Answer 23

It's just five haha...

Answer 24

And yes, I'm happy with it. I needed to make sure that I wasn't going completely insane. Thank you :)

Answer 25

glad to help