Question

PLEASE HELP :( !!
y=13e^(3x^2)-6x
at x=3 in the form y=mx +b has

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

omg this is so simple

Answer 5

wat is america coming to

Answer 6

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

Answer 7

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

Answer 8

@akotto4897 wtf is wrong with you? Why would you call someone elses problem `so simple`.

It's like someone in 5th grade calling 3rd grade math SUPER EASY!
Just because you understand it doesn't mean that everyone else is on your level...

Answer 9

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

Answer 10

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

Answer 11

Mmm ok so your slope looks good.
So what coordinate pair are we using to find our y-intercept?\[\Large f(3)\quad=\quad 13e^{27}-18\]Mmm ok ok I see you have the written down already :3

\[\Large y_{\tan}\quad=\quad mx+b\]So if we plug in our coordinate pair, we get something like this, yes?\[\Large 13e^{27}-18\quad=\quad (234e^{27}-6)\cdot 3+b\]

Answer 12

woah so the slope is good and the y too.

Answer 13

oh i see there's an x there.

Answer 14

so would it be 247e^27-18?

Answer 15

for your y_tan? or for b?

Answer 16

y tan

Answer 17

Hmm after you find your b value, you should plug it back into the \[\Large\bf y_{tan}\quad=\quad \color{royalblue}{m}x+\color{royalblue}{b}\]
For our final answer, we DON'T want the coordinate pair plugged in.
We're only trying to fill in these blue pieces.

Answer 18

so how do i get to the answer.

Answer 19

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

Answer 20

grr sorry website crashed on me >:c

Answer 21

its fine

Answer 22

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

Answer 23

i keep getting the answer wrong.

Answer 24

\[\Large\bf y_{\tan}\quad=\quad \color{royalblue}{m}x+\color{royalblue}{b}\]Plugging in our slope,\[\Large\bf y_{\tan}\quad=\quad \color{royalblue}{(234e^{27}-6)}x+\color{royalblue}{b}\]Then we plug in our coordinate pair to find b,\[\Large\bf \color{#DD4747 }{13e^{27}-18}\quad=\quad \color{royalblue}{(234e^{27}-6)}\cdot\color{#DD4747 }{3}+\color{royalblue}{b}\]

Answer 25

So we have some -18's cancelling out.
And I think we get a b value of,\[\Large\bf \color{royalblue}{b\quad=\quad -689e^{27}}\]Something like that?

Answer 26

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

Answer 27

y-?

Answer 28

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

Answer 29

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

Answer 30

and x1= 3

Answer 31

No that's `point-slope` form of a line.
y-y1 = m( x-x1)
We don't want to use that.

We were told to use the `slope-intercept` form of a line.
y=mx+b

Answer 32

can we do it the way i had it please?

Answer 33

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

Answer 34

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

Answer 35

Yes we can use that, so let's see if we can get it set up correctly.
Lemme see if I can match what you wrote a sec :)

Answer 36

ok

Answer 37

\[\Large\bf y-y_1\quad=\quad \color{royalblue}{m}(x-x_1)\]\[\Large\bf y-y_1\quad=\quad \color{royalblue}{(234e^{27}-6)}(x-x_1)\]Then plugging in our point:\[\Large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad \color{royalblue}{(234e^{27}-6)}(x-\color{#DD4747 }{3})\]

Answer 38

yes correct! :)

Answer 39

after that I'm lost lol

Answer 40

Ok so now we just need to get in into the form y=mx+b.
So we need to multiply out the brackets, then we gotta isolate the y term.

Answer 41

\[\large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad 234e^{27}x-6x -702e^{27}+18\]I think it expands like that, yes?
Give it a try :)

Answer 42

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

Answer 43

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

Answer 44

and leave the

Answer 45

Yes, true. We probably want don't want the x's expanded like that,\[\large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad (234e^{27}-6)x -702e^{27}+18\]

Answer 46

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

Answer 47

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

Answer 48

and then -3

Answer 49

\[\large\bf (234e^{27}-6)(x-3)\quad=\\ \large\bf\quad (234e^{27}-6)x-(234e^{27}-6)3\]And we only want to multiply out the part with the 3.

Answer 50

im so confused. i do't understand why we did this problem differently than all the others when it comes to this part.

Answer 51

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

Answer 52

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

Answer 53

Are you not getting that when you foil..?

Answer 54

yes i have that! :)

Answer 55

We don't want multiple x terms, so we'll factor an x out of each of the first two terms.\[\Large\bf =\quad \large\bf (234e^{27}-6)x -702e^{27}+18\]

Answer 56

now what? :(

Answer 57

where does my foiling go?

Answer 58

great so we factor?

Answer 59

We had this,\[\Large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad \color{royalblue}{(234e^{27}-6)}(x-\color{#DD4747 }{3})\]It foiled to give us this,\[\large\bf y-(13e^{27}-18)\quad=\quad 234e^{27}x-6x -702e^{27}+18\]We factored out an x to give us this,\[\large\bf y-(13e^{27}-18)\quad=\quad (234e^{27}-6)x -702e^{27}+18\]

Answer 60

We finish by solving for y.

Answer 61

right

Answer 62

x(-468+12)

Answer 63

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

Answer 64

what..? :(

Answer 65

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

Answer 66

I don't know why you insisted on using point-slope form for this one.
It seems way more complicated :c

Answer 67

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

Answer 68

This one only ended up being complicated because our slope was `2 terms`.
So it ended up being a ton of multiplication.

Answer 69

damn im so confused.

Answer 70

=[

Answer 71

what would be he answer btw?

Answer 72

\[\large\bf y-(13e^{27}-18)\quad=\quad (234e^{27}-6)x -702e^{27}+18\]From here: you add (13e^{27}-18) to each side,\[\large\bf y=\\ \large\bf(234e^{27}-6)x -702e^{27}+18\color{orangered}{+(13e^{27}-18)}\]Does that step make sense? +_+
That's how we isolate the y term.

Answer 73

From there we can drop the brackets since nothing in being applied to them,
\[\large\bf y=\\ \large\bf(234e^{27}-6)x -702e^{27}+18+13e^{27}-18\]It looks like the 18's will cancel out,\[\large\bf y=\\ \large\bf(234e^{27}-6)x -702e^{27}+\cancel{18}+13e^{27}-\cancel{18}\]

Answer 74

And the exponential terms ( not the one attached to the x ) are like terms, so we'll combine them,\[\large\bf y=\\ \large\bf(234e^{27}-6)x \color{orangered}{-702e^{27}+13e^{27}}\]
\[\large\bf y=(234e^{27}-6)x \color{orangered}{-689e^{27}}\]And I think that's our final answer, assuming I didn't make any boo boos along the way.

Answer 75

wait which is the mx and b?

Answer 76

oh it is correct….

Answer 77

\[\large\bf y=\color{royalblue}{(234e^{27}-6)}x \color{orangered}{-689e^{27}}\]\[\large\bf y=\color{royalblue}{m}x +\color{orangered}{b}\]