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

- anonymous

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

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- amistre64

what are 2 things we need to know to define a line?

- anonymous

two points?

- amistre64

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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## More answers

- anonymous

slope?

- amistre64

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?

- anonymous

the change in y over the change in x? so the point slope equation?

- amistre64

hmm, im assuming you know what a derivative is for this to work.

- anonymous

yeah but I don't really get derivatives

- anonymous

I know there's 4 types

- amistre64

you should have a simple rule you can apply
constant rule, and a power rule

- amistre64

\[\frac{d}{dx}k=0\]
\[\frac{d}{dx}x^n=n~x^{n-1}\]

- anonymous

I'm just having a hard time applying the equation to these rules

- amistre64

9 - x^2
^ ^
^ this is a power of x
^
and this is a constant

- amistre64

what is the derivative of 9?
what is the derivative of -x^2?

- amistre64

e3rr, -7x^2 ... forgot the equation we were dealing with :/

- anonymous

so for 2 it would be 0 and -7x^2 would be -7x^-6?

- amistre64

\[\frac{d}{dx}(2-7x^2)\]
\[\frac{d}{dx}(29)-\frac{d}{dx}(7x^2)\]
\[\frac{d}{dx}(2)-7\frac{d}{dx}(x^2)\]
\[0-7(2x^{2-1})\]

- anonymous

how did you get 29?

- amistre64

typo, slip of the fingers.

- anonymous

can you please explain how you got the answer step by step?

- amistre64

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.

- amistre64

your -7x^2 derivative is off ... try it again

- amistre64

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

- anonymous

sorry my book shows different explanations so I'm a little confused right now

- amistre64

tell me your thoughts then, what does your book explain?

- anonymous

it basically shows me all these rules but all these variables are confusing me

- anonymous

but the examples only show me if I have a point and slope

- amistre64

what is your rule for powers of x?

- amistre64

can you take a picture of it?

- anonymous

For any real number k if y=x^k then d/dx(x^h)= k*x^k-1

- amistre64

let k=2
x^2 become 2x^(2-1)
or simply 2x

- amistre64

so our derivative is
-7x^2 derives to -7(2x), at x=2 that is ... -7(4)

- anonymous

so the 2 becomes x?

- amistre64

out point (2,-26) defines the value of x as 2

- amistre64

*our point ...

- amistre64

2x, when x=2 is just ... 2(2) right?

- anonymous

how did x^2 become 2x?

- amistre64

by the derivative rule you posted. let k=2

- amistre64

x^k derives to k*x^(k-1)

- anonymous

Ohhhh I get it now

- amistre64

by first principles of limit
\[\lim_{h\to0}\frac{(x+h)^2-x^2}{(x+h)-x}\]
\[\lim_{h\to0}\frac{x^2+h^2+2xh-x^2}{h}\]
\[\lim_{h\to0}\frac{h^2+2xh}{h}\]
\[\lim_{h\to0}\frac{h(h+2x)}{h}\]
\[\lim_{h\to0}(h+2x)=2x\]

- anonymous

but I'm still pretty confused

- amistre64

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.

- anonymous

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?

- amistre64

well, thats our slope of our tangent line. the rest is from algebra class.

- amistre64

construct the line using your slope and the stated point

- anonymous

y=mx+b

- anonymous

or point slope equation
y-y1=m(x-x1)

- amistre64

y - f(a) = f'(a) (x-a)
or rewritten as
y = f'(a) (x-a) +f(a)

- amistre64

point slope equation

- anonymous

y-(-26)=-28(x-2)

- amistre64

thatll do it

- anonymous

y=-28x+30

- anonymous

could I get a two out of it to be 2(-14x+15)? or is okay like that?

- amistre64

usually we dont factor it, keep it in y=mx+b form is conventional

- anonymous

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

- amistre64

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.

- anonymous

okay thank you!!

- amistre64

youre welcome :) and good luck

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