anonymous
  • anonymous
help! Given determine the points where (a) the curve intercepts the axis; (b) the tangent line to the curve is horizontal.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
i believe that the 2nd answer is (5,25) right? @dlxhazedxlb @hihihii
anonymous
  • anonymous
yup! 2nd answer is (5,25) good job!

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More answers

anonymous
  • anonymous
how about the first?
anonymous
  • anonymous
im not sure how to do this one
anonymous
  • anonymous
@hihihii
anonymous
  • anonymous
in order to find where the curve intersects the x-axis, you have to set the original function equal to 0. -1x^2+ 10x= 0
anonymous
  • anonymous
x^2-10x=0 x(x-10)=0 x=0, x=10 to find the points, plug 0 and 10 back into the original equation
anonymous
  • anonymous
0+0=0 (0,0) -1(10)^2-10(10)=0 (10,0) both the y values are 0 because that's where it crosses the x-axis
anonymous
  • anonymous
@hihihii i see where you got x=0 but how'd you get -10?
anonymous
  • anonymous
i got -x+10
anonymous
  • anonymous
and also which is the correct answer to A?
anonymous
  • anonymous
(0,0) or (10,0)? which one do i pick?
anonymous
  • anonymous
@hihihii
anonymous
  • anonymous
x^2-10x=0 ??
anonymous
  • anonymous
hello hello hello
anonymous
  • anonymous
@satellite73 @SithsAndGiggles would anyone kindly like to help?
anonymous
  • anonymous
\[f(x)=-x^2+10x\] Finding where the curve intercepts the x-axis is the same as solving for the roots of the equation: \[-x^2+10x=0\] The tangent line to the curve is horizontal when the slope of the tangent line is 0. Hence solve for x in \(f'(x)=0\): \[f'(x)=-2x+10=0\]

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