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mathcalculus
 one year ago
help! Given determine the points where
(a) the curve intercepts the axis;
(b) the tangent line to the curve is horizontal.
mathcalculus
 one year ago
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
 one year ago
Best ResponseYou've already chosen the best response.0i believe that the 2nd answer is (5,25) right? @dlxhazedxlb @hihihii

hihihii
 one year ago
Best ResponseYou've already chosen the best response.1yup! 2nd answer is (5,25) good job!

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0how about the first?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0im not sure how to do this one

hihihii
 one year ago
Best ResponseYou've already chosen the best response.1in order to find where the curve intersects the xaxis, you have to set the original function equal to 0. 1x^2+ 10x= 0

hihihii
 one year ago
Best ResponseYou've already chosen the best response.1x^210x=0 x(x10)=0 x=0, x=10 to find the points, plug 0 and 10 back into the original equation

hihihii
 one year ago
Best ResponseYou've already chosen the best response.10+0=0 (0,0) 1(10)^210(10)=0 (10,0) both the y values are 0 because that's where it crosses the xaxis

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0@hihihii i see where you got x=0 but how'd you get 10?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0and also which is the correct answer to A?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0(0,0) or (10,0)? which one do i pick?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0hello hello hello

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 @SithsAndGiggles would anyone kindly like to help?

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x)=x^2+10x\] Finding where the curve intercepts the xaxis 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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