Find the tangent line to y^2=x at the point (4,-2).
I know that dy/dx= 1/(2y) and I know how to write the tangent line function. But since there is no x value to find the slope, would the function not exists?

Question

Find the tangent line to y^2=x at the point (4,-2). 
I know that dy/dx= 1/(2y) and I know how to write the tangent line function. But since there is no x value to find the slope, would the function not exists?

davejavous · Answer

Use implicit differentiation on y^2=x

canada907cat · Answer

i did use implicit differentiation. That's how I got 1/(2y)

danjs · Answer

you can re-write the equation for y

$$\large y = x ^{1/2}$$
$$\frac{ dy }{ dx }=\frac{ 1 }{ 2\sqrt{x} }$$

canada907cat · Answer

okay so how would I solve it from there in terms of putting it into the tangent function?

danjs · Answer

$$y = \pm \sqrt{x}$$

danjs · Answer

the point (x , y) = (4 , -2) is given to use

The slope of a tangent here at x=4 is
$$\frac{ dy }{ dx }=\frac{ 1 }{ 2\sqrt{4} }=\frac{ 1 }{ 4 }$$

canada907cat · Answer

okay so I would put 1/4 into the tangent formula correct?

danjs · Answer

sorry you need the negative slope for that point 
$$\frac{ dy }{ dx }=-1/4$$

danjs · Answer

When x=4, y=-2,  that is the negative root

canada907cat · Answer

Okay so would I do anything with 1/(2y) or would I just put in the -1/4 for the slope in the tangent function. Oh okay.

danjs · Answer

yeah so you have the slope and a point the line goes through..

line slope = -1/4 through point (4 , -2)

y - (-2) = (-1/4)*(x - 4)

canada907cat · Answer

Don't you mean T(x)= -2-1/4(x-4)?

danjs · Answer

you can clean it up a bit..

y - (-2) = (-1/4)*(x - 4)
$$y = \frac{ -1 }{ 4 }x-1$$