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f'(x)= 78xe^(3x^2)-6

slope= 234e^27-6
y=13e^27-18

So you're looking for an equation of the line tangent to the curve f(x) at x=3?

omg this is so simple

wat is america coming to

yes, Determine the equation of the line tangent to the graph of

yeah but I'm not sure why I'm getting this wrong.

Ok that's enough of my rant :) lemme check your work a sec :3

lol, i appreciate that. but yes, let's begin.

woah so the slope is good and the y too.

oh i see there's an x there.

so would it be 247e^27-18?

for your y_tan? or for b?

y tan

so how do i get to the answer.

at first i thought it was 247e^27 -6x

grr sorry website crashed on me >:c

its fine

i just don't understand what to do at the point slope formula.

i keep getting the answer wrong.

isn't it y-(13e^27-18)=234e^27-6x+18

y-?

right, y-y1=m(x-x1)

so in this case, slope= 234e^27-6
y=13e^27-18

and x1= 3

can we do it the way i had it please?

yeah but i thought i could use that to find the equation of a tangent line

I've been using that for other problems and it worked.

ok

yes correct! :)

after that I'm lost lol

yup, but i seem to be doing that wrong unfortunately.

wait don't we just multiply out the -6 and -6*-3 only?

and leave the

\[\Large 234e^{27}\cdot-3\]

why don't we multiply 234e^27 with x?

and then -3

so for (234e^27−6)(x−3) i thought we do foil

Yes, foiling should give you this,\[\large\bf 234e^{27}x-6x -702e^{27}+18\]

Are you not getting that when you foil..?

yes i have that! :)

now what? :(

where does my foiling go?

great so we factor?

We finish by solving for y.

right

x(-468+12)

\[\large\bf y=\\\large\bf (234e^{27}-6)x -702e^{27}+18+13e^{27}-18\]

what..? :(

the -6 and the 18 are not like terms, we can't combine those.

but it's the only way for me… i have to stick to this or else i'll be completely lost in the exam

damn im so confused.

=[

what would be he answer btw?

wait which is the mx and b?

oh it is correct….