Saturday

January 31, 2015

January 31, 2015

Posted by **anonymous** on Monday, January 28, 2008 at 6:38pm.

find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of the points.

- calculus -
**Reiny**, Monday, January 28, 2008 at 9:02pmsub x = 2 into your equation to get

4 - 4y = y^2 - 1

0 = y^2 + 4y - 5

(y+5)(y-1) = 0

so when x=2, y =-5 or when x=2, y = 1

so the points are (2,-5) and (2,1)

check them by subbing them back into the original equation, they work

- calculus -
**Reiny**, Monday, January 28, 2008 at 9:38pmI found the derivative of your equation implicitly and got

dy/dx = (2x+2xy)/(2y = x^2)

for a horizontal tangent this has to be zero, which means the numerator should be zero

subbing in (2,-5) makes the top zero, so there is a horizontal tangent at the point (2,_5)

**Answer this Question**

**Related Questions**

Calculus - Consider the curve y^2+xy+x^2=15. What is dy/dx? Find the two points ...

calculus - dy/dx= (2x-2xy)/(x^2+2y) find all points on the curve where x=2. show...

calculus - dy/dx= (2x-2xy)/(x^2+2y) Find all points on the curve where x=2. Show...

calculus - dy/dx= (2x-2xy)/(x^2+2y) Find all points on the curve where x=2. Show...

Calculus - Consider the curve given by y^2 = 2+xy (a) show that dy/dx= y/(2y-x...

calculus - Given the curve defined by the equation y=cos^2(x) + sqrt(2)* sin(x) ...

Calculus - Please help this is due tomorrow and I dont know how to Ive missed a ...

Calculus - a)The curve with equation: 2y^3 + y^2 - y^5 = x^4 - 2x^3 + x^2 has ...

calculus - 1. Consider the function f(x) = X^(4/3) +4x^(1/3) on the interval -8...

calculus - 1. Consider the function f(x) = X^(4/3) +4x^(1/3) on the interval -8...