mathcalculus
help! Given determine the points where
(a) the curve intercepts the axis;
(b) the tangent line to the curve is horizontal.
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mathcalculus
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i believe that the 2nd answer is (5,25) right? @dlxhazedxlb @hihihii
hihihii
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yup! 2nd answer is (5,25) good job!
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how about the first?
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im not sure how to do this one
mathcalculus
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@hihihii
hihihii
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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
hihihii
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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
hihihii
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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
mathcalculus
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@hihihii i see where you got x=0 but how'd you get -10?
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i got -x+10
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and also which is the correct answer to A?
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(0,0) or (10,0)? which one do i pick?
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@hihihii
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x^2-10x=0 ??
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hello hello hello
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@satellite73 @SithsAndGiggles would anyone kindly like to help?
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\[f(x)=-x^2+10x\]
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 \(f'(x)=0\):
\[f'(x)=-2x+10=0\]