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help! Given determine the points where (a) the curve intercepts the axis; (b) the tangent line to the curve is horizontal.

Mathematics
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i believe that the 2nd answer is (5,25) right? @dlxhazedxlb @hihihii
yup! 2nd answer is (5,25) good job!

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Other answers:

how about the first?
im not sure how to do this one
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
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
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
@hihihii i see where you got x=0 but how'd you get -10?
i got -x+10
and also which is the correct answer to A?
(0,0) or (10,0)? which one do i pick?
x^2-10x=0 ??
hello hello hello
@satellite73 @SithsAndGiggles would anyone kindly like to help?
\[f(x)=-x^2+10x\] 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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