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f '(x)=4x^3+6x did you get this far?
ok i know how you got there
10 is the slope of the tangent line at x=-1
so we know a point on the line and we know the slope of that line and a line has form y=mx+b f(-1)=10(-1)+b we can find b now 1+3=-10+b 4=-10+b 10+4=b 14=b so the tangent line is y=10x+14
For whatever reason the answers in the back of the book are telling me the tangent would equal -10x-6, is it possible we mixed signs along the way?
let me do it on paper one sec
is this what you are confused about? the answer in the back of the book? did you try this problem and get what I got?
oh i plug in 1
You didn't plug in -1
No i got something like x^5+5x^4+7x^3+9x^2+6x because i thought it was F(x)=f(a)+f'(a)(x-a) inorder to solve this beast
f '(-1)=-10 not 10
y=-10x+b we need to know b we know a point on this line
Nah, just find the slope of the tangent and plug in that slope into point slope formula
Ok after you get the slope of the tangent what did you plug in to get you (x,y) values?
If x=1, then y=....
damn stupid 1
\(x = -1 \implies f = x^4+3x^2 = ?\)
thats 4 but we are stillnot jiving with the stupid books answer
we are fixing to
so we have y=-10x+b 4=-10(-1)+b
Take the slope (-10) and the point (-1,4) and plug into point slope. \(y-y_0 = m(x-x_0)\)
solve for b
you can use either way
Where m is the slope, \(x_0,y_0\) are the x and y for your point.
b=4-10=-6 so y=-10x-6
I like point slope cause I don't have to do anything else. Slope intercept is only useful if you're gonna graph it.
i like slope intercept its more fun to me for some reason
Ok where did we pull the y value from (4) sorry i missed how we came up with that one
Plug in x=-1 to the original equation for f(x)
Nevermind I'm sorry, I just figured that part out, thanks for the help guys. It has been a long day to put it nicely.