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

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mathcalculusBest ResponseYou've already chosen the best response.0
i believe that the 2nd answer is (5,25) right? @dlxhazedxlb @hihihii
 8 months ago

hihihiiBest ResponseYou've already chosen the best response.1
yup! 2nd answer is (5,25) good job!
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
how about the first?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
im not sure how to do this one
 8 months ago

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

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

hihihiiBest ResponseYou've already chosen the best response.1
0+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
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@hihihii i see where you got x=0 but how'd you get 10?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
and also which is the correct answer to A?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
(0,0) or (10,0)? which one do i pick?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
hello hello hello
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@satellite73 @SithsAndGiggles would anyone kindly like to help?
 8 months ago

SithsAndGigglesBest 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 \(f'(x)=0\): \[f'(x)=2x+10=0\]
 8 months ago
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