anonymous
  • anonymous
Find an equation of the tangent line to the graph f(x)=2-7x^2 at (2,-26)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
what are 2 things we need to know to define a line?
anonymous
  • anonymous
two points?
amistre64
  • 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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anonymous
  • anonymous
slope?
amistre64
  • 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
  • anonymous
the change in y over the change in x? so the point slope equation?
amistre64
  • amistre64
hmm, im assuming you know what a derivative is for this to work.
anonymous
  • anonymous
yeah but I don't really get derivatives
anonymous
  • anonymous
I know there's 4 types
amistre64
  • amistre64
you should have a simple rule you can apply constant rule, and a power rule
amistre64
  • amistre64
\[\frac{d}{dx}k=0\] \[\frac{d}{dx}x^n=n~x^{n-1}\]
anonymous
  • anonymous
I'm just having a hard time applying the equation to these rules
amistre64
  • amistre64
9 - x^2 ^ ^ ^ this is a power of x ^ and this is a constant
amistre64
  • amistre64
what is the derivative of 9? what is the derivative of -x^2?
amistre64
  • amistre64
e3rr, -7x^2 ... forgot the equation we were dealing with :/
anonymous
  • anonymous
so for 2 it would be 0 and -7x^2 would be -7x^-6?
amistre64
  • 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
  • anonymous
how did you get 29?
amistre64
  • amistre64
typo, slip of the fingers.
anonymous
  • anonymous
can you please explain how you got the answer step by step?
amistre64
  • 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
  • amistre64
your -7x^2 derivative is off ... try it again
amistre64
  • 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
  • anonymous
sorry my book shows different explanations so I'm a little confused right now
amistre64
  • amistre64
tell me your thoughts then, what does your book explain?
anonymous
  • anonymous
it basically shows me all these rules but all these variables are confusing me
anonymous
  • anonymous
but the examples only show me if I have a point and slope
amistre64
  • amistre64
what is your rule for powers of x?
amistre64
  • amistre64
can you take a picture of it?
anonymous
  • anonymous
For any real number k if y=x^k then d/dx(x^h)= k*x^k-1
amistre64
  • amistre64
let k=2 x^2 become 2x^(2-1) or simply 2x
amistre64
  • amistre64
so our derivative is -7x^2 derives to -7(2x), at x=2 that is ... -7(4)
anonymous
  • anonymous
so the 2 becomes x?
amistre64
  • amistre64
out point (2,-26) defines the value of x as 2
amistre64
  • amistre64
*our point ...
amistre64
  • amistre64
2x, when x=2 is just ... 2(2) right?
anonymous
  • anonymous
how did x^2 become 2x?
amistre64
  • amistre64
by the derivative rule you posted. let k=2
amistre64
  • amistre64
x^k derives to k*x^(k-1)
anonymous
  • anonymous
Ohhhh I get it now
amistre64
  • 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
  • anonymous
but I'm still pretty confused
amistre64
  • 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
  • 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
  • amistre64
well, thats our slope of our tangent line. the rest is from algebra class.
amistre64
  • amistre64
construct the line using your slope and the stated point
anonymous
  • anonymous
y=mx+b
anonymous
  • anonymous
or point slope equation y-y1=m(x-x1)
amistre64
  • amistre64
y - f(a) = f'(a) (x-a) or rewritten as y = f'(a) (x-a) +f(a)
amistre64
  • amistre64
point slope equation
anonymous
  • anonymous
y-(-26)=-28(x-2)
amistre64
  • amistre64
thatll do it
anonymous
  • anonymous
y=-28x+30
anonymous
  • anonymous
could I get a two out of it to be 2(-14x+15)? or is okay like that?
amistre64
  • amistre64
usually we dont factor it, keep it in y=mx+b form is conventional
anonymous
  • 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
  • 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
  • anonymous
okay thank you!!
amistre64
  • amistre64
youre welcome :) and good luck

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