Question

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

Answer 1

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

Answer 2

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

Answer 3

how about the first?

Answer 4

im not sure how to do this one

Answer 5

@hihihii

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

i got -x+10

Answer 11

and also which is the correct answer to A?

Answer 12

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

Answer 13

@hihihii

Answer 14

x^2-10x=0 ??

Answer 15

hello hello hello

Answer 16

@satellite73 @SithsAndGiggles would anyone kindly like to help?

Answer 17

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