mathcalculus 2 years ago help! Given determine the points where (a) the curve intercepts the axis; (b) the tangent line to the curve is horizontal.

1. mathcalculus

2. mathcalculus

i believe that the 2nd answer is (5,25) right? @dlxhazedxlb @hihihii

3. hihihii

yup! 2nd answer is (5,25) good job!

4. mathcalculus

5. mathcalculus

im not sure how to do this one

6. mathcalculus

@hihihii

7. hihihii

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

8. hihihii

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

9. hihihii

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

10. mathcalculus

@hihihii i see where you got x=0 but how'd you get -10?

11. mathcalculus

i got -x+10

12. mathcalculus

and also which is the correct answer to A?

13. mathcalculus

(0,0) or (10,0)? which one do i pick?

14. mathcalculus

@hihihii

15. mathcalculus

x^2-10x=0 ??

16. mathcalculus

hello hello hello

17. mathcalculus

@satellite73 @SithsAndGiggles would anyone kindly like to help?

18. SithsAndGiggles

\[f(x)=-x^2+10x\] 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 \(f'(x)=0\): \[f'(x)=-2x+10=0\]

Find more explanations on OpenStudy