Find an equation of the tangent line to the graph f(x)=2-7x^2 at (2,-26)

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Find an equation of the tangent line to the graph f(x)=2-7x^2 at (2,-26)

Mathematics
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what are 2 things we need to know to define a line?
two points?
well, 2 points are useful yes, but we only have one point given to us ... what else could we use to define a line with? a point and ....

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slope?
correct, so a point and a slope can be used to define a line. its called the point slope form of the equation. now, how do we determine the slope of a curve at a given point?
the change in y over the change in x? so the point slope equation?
hmm, im assuming you know what a derivative is for this to work.
yeah but I don't really get derivatives
I know there's 4 types
you should have a simple rule you can apply constant rule, and a power rule
\[\frac{d}{dx}k=0\] \[\frac{d}{dx}x^n=n~x^{n-1}\]
I'm just having a hard time applying the equation to these rules
9 - x^2 ^ ^ ^ this is a power of x ^ and this is a constant
what is the derivative of 9? what is the derivative of -x^2?
e3rr, -7x^2 ... forgot the equation we were dealing with :/
so for 2 it would be 0 and -7x^2 would be -7x^-6?
\[\frac{d}{dx}(2-7x^2)\] \[\frac{d}{dx}(29)-\frac{d}{dx}(7x^2)\] \[\frac{d}{dx}(2)-7\frac{d}{dx}(x^2)\] \[0-7(2x^{2-1})\]
how did you get 29?
typo, slip of the fingers.
can you please explain how you got the answer step by step?
i jsut did ... except for 29 which is spose to be a 2 instead use the constant rule for the constant, and the power rule for the power of x there is also a 'pull out' rule which lets us pull out that 7 to the outside.
your -7x^2 derivative is off ... try it again
the format for the tangent line is just gonna be: y - f(a) = f'(a) (x-a) so as long as we can determine f'(a) we are set
sorry my book shows different explanations so I'm a little confused right now
tell me your thoughts then, what does your book explain?
it basically shows me all these rules but all these variables are confusing me
but the examples only show me if I have a point and slope
what is your rule for powers of x?
can you take a picture of it?
For any real number k if y=x^k then d/dx(x^h)= k*x^k-1
let k=2 x^2 become 2x^(2-1) or simply 2x
so our derivative is -7x^2 derives to -7(2x), at x=2 that is ... -7(4)
so the 2 becomes x?
out point (2,-26) defines the value of x as 2
*our point ...
2x, when x=2 is just ... 2(2) right?
how did x^2 become 2x?
by the derivative rule you posted. let k=2
x^k derives to k*x^(k-1)
Ohhhh I get it now
by first principles of limit \[\lim_{h\to0}\frac{(x+h)^2-x^2}{(x+h)-x}\] \[\lim_{h\to0}\frac{x^2+h^2+2xh-x^2}{h}\] \[\lim_{h\to0}\frac{h^2+2xh}{h}\] \[\lim_{h\to0}\frac{h(h+2x)}{h}\] \[\lim_{h\to0}(h+2x)=2x\]
but I'm still pretty confused
the 'rules' are really a table of outcomes that can be applied so that you dont have to use the long limiting process each and every time.
so now that we know the derivative of -7x^2 to be -7(2x) which is -7(2*2)= -28 what do we do next?
well, thats our slope of our tangent line. the rest is from algebra class.
construct the line using your slope and the stated point
y=mx+b
or point slope equation y-y1=m(x-x1)
y - f(a) = f'(a) (x-a) or rewritten as y = f'(a) (x-a) +f(a)
point slope equation
y-(-26)=-28(x-2)
thatll do it
y=-28x+30
could I get a two out of it to be 2(-14x+15)? or is okay like that?
usually we dont factor it, keep it in y=mx+b form is conventional
so that's my final answer? thank you! is there any way I reward you on here? it's my first time using this site
there should be a 'best response' button to everything ive posted. clicking it gives out a medal. but saying thank you and learning a little bit is reward enough for me to be honest.
okay thank you!!
youre welcome :) and good luck

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