Posted by anonymous on Monday, January 28, 2008 at 6:38pm.
sub 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
I 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)
Related Questions
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 y^2+xy+x^2=15. What is dy/dx? Find the two points ...
Calculus - a)The curve with equation: 2y^3 + y^2 - y^5 = x^4 - 2x^3 + x^2 has ...
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 - 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...
Calculus - the curve: (x)(y^2)-(x^3)(y)=6 (dy/dx)=(3(x^2)y-(y^2))/(2xy-(x^3)) a...
For Further Reading
