anonymous
  • anonymous
A. Find the derivative of f at x. That is, find f'(x). B. Find the slope of the tangent line to the graph of f at each of the two values of x given to the right of the function f(x)=x^2-8; x=-1, x=3
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
misty1212
  • misty1212
HI!!
anonymous
  • anonymous
Hey

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misty1212
  • misty1212
you know the derivative of \(x^2\)?
misty1212
  • misty1212
if the answer is "no" that is fine, i will tell you
anonymous
  • anonymous
is it 2x?
misty1212
  • misty1212
yes
misty1212
  • misty1212
and the derivative of a constant is zero, making the derivative of \(f(x)=x^2-8\) the function \[f'(x)=2x\] that is all
anonymous
  • anonymous
but its two parts to the question is 2x the answer to them both
anonymous
  • anonymous
misty1212
  • misty1212
no \(2x\) is the derivative you now need the equation for two lines, one where \(x=-1\) and the other where \(x=-3\)
anonymous
  • anonymous
f(x)=-1^2-8 and f(x)=-3^2-8
misty1212
  • misty1212
if \(x=-1\) then \(y=(-1)^2-8=-7\) so the point is \((-1,-7)\) and the slope is \(2\times -1=-2\) use the point slope formula to get the equation of the line
misty1212
  • misty1212
if \(x=3\) then \(y=3^2-8=1\) the point is \((3,1)\) and the slope is \(2\times 3=6\) points slope again
anonymous
  • anonymous
y-(-7)=-2(x-(-1)) and y-(1)=6(x-(3))
anonymous
  • anonymous
misty1212
  • misty1212
you can clean them up a great deal, but yes
anonymous
  • anonymous
y=-2x+5 y=6x-17
anonymous
  • anonymous
anonymous
  • anonymous
o Thank you

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