Find the equations of the tangent lines to the graph of y2-xy+3=0 at= -4?

Question

help me, pls

Answer 1

Do you mean at x = -4?

Firstly, find the y values. We have y^2 + 4y + 3 = 0, (y+1)(y+3) = 0, so y = -1 or -3.

Now, the next thing you need to do is find the gradient. Differentiate implicitly:
y^2 differentiates to 2yy', and xy differentiates to y + xy' by the chain rule.
So we get:
2y * y' - (y + xy') = 0
y'(2y - x) = y
y' = y/(2y-x).

For the first point (-4,-1), we get y' = -0.5.
For the second point (-4,-3), we get y' = 1.5.

Thus, using y - y1 = m(x-x1), we get:
y + 1 = -0.5(x+4), or y = -0.5x - 3 for the first tangent.
y + 3 = 1.5(x+4) or y = 1.5x + 3 for the second tangent.

Answer 2

_______________________________ ??????? (2x+y)? - 3y + xy - 4 = 0 Differentiating implicitly with respect to x: 4(2x+y)³·(2 + y') - 3y' + [ xy' + y·a million ] - 0 = 0 ???? ? Now, replace 2 and -3 for x and y ?? 4(2·2-3)³·(2 + y') - 3y' + [ 2y' - 3 ] - 0 = 0 ?????????? ?????????? 4(2 + y') - 3y' + 2y' - 3 = 0 ?????????? ?????????? ?? 8 + 4y' - 3y' + 2y' - 3 = 0 ?????????? ?????????? ?????????? ?????????? ??? 3y' = ??-5 ?????????? ?????????? ?????????? ?????????? ????? y' = ??-5/3 Now, using the point-slope form of a line for element (2,-3) and slope m = y' = ??-5/3, we get that the equation of the tangent line at (2,-3) is: ?????????? ?????? y - (-3) = (-5/3)(x - 2) ?????????? ?????????? ?? ????y = ?-(5/3)x + a million/3 ?????????? ? answer ________________________________

Answer 3

Yeah for calculus!
tangent lines can be found using the derivatives of graphs.
the derivative of y is y prime or y to the 1st (y^1)
first take the derivative which is:
(y^1)y - y - x(y^1) = 0
get all y prime terms on one side
(y^1)y - x(y^1) = y
factor out y prime
(y^1)(y-x) = y
solve for y prime
y^1 = y/(y-x) which reduces to y^1 = -1/x
-1/x is the slope of your tangent line, so at -4 (plug in for x) the slope is 1/4
use point slope form
y - ... = (1/4)(x - (-4))
to get the ... plug -4 into the original equation and solve for y which gives you both -4.646 and 0.646 so each of them is a tangent line at x= -4. So your final answers are
y + 4.646 = (1/4)(x+4)
and
y - 0.646 = (1/4)(x+4)

Answer 4

I'm going to assume you mean at x= -4.
First you need to find the y values that correspond to x=-4
plugging in -4 for x gives
y^2+4y+3=0
(y+3)(y+1)=0
y=-3 or y=-1
So now we have two points (-4,-3) and (-4,-1) at which we need the slope of the tangent line.
We will implicitly differentiate the equation.
2y(dy/dx)-(x(dy/dx)+y(1))=0
2y(dy/dx)-x(dy/dx)-y=0
(2y-x)(dy/dx)=y
dy/dx=y/(2y-x)
Now plugging in at each point above gives the slopes of each tangent lin
at (-4,-3), dy/dx=-3/(2(-3)-(-4))=3/2
at (-4,-1), dy/dx=-1/(2(-1)-(-4))=-1/2
Using the point slope equation for a line:

y-(-3)=(3/2)(x-(-4)), y+3=(3/2)x + 6, y=(3/2)x +3
and
y-(-1)=(-1/2)(x-(-4)), y+1=(-1/2)x -2, y=(-1/2)x -3

Those are the equations you want

Answer 5

Eh, I screwed up....

Go with Stephen's answer.

--------------------------

Implicit differentiation.

y² - xy + 3 = 0
2yy' - x = 0
2yy' = x
y' = x/(2y)

at x=-4,

y² +4y + 3 = 0
(y+3)(y+1) = 0

So y is either -3 or -1...there are 2 tangent lines!

if y=-3 then y' = -4/[(2)(-3)] = 2/3

(y + 3) / (x + 4) = 2/3
y+3 = 2/3x + 8/3
y = -2/3x - 1/3

If y=-1 then y' = -4/[2(-1)] = 2
(y + 1) / (x + 4) = 2
y+1 = 2x + 8
y = 2x + 7

Answer 6

First, take the derivative, which is 2y(y') -(y+x(y')) +0 = 0. Then, gather all the y' s on one side of the equation: 2y(y') - x(y') = y. Then, y' (2y - x) = y, and y' = y/(2y-x). Then to solve at (I'm assuming you mean...) x = -4, just plug in that x-value to get y' = y/(2y+4). I don't know how your teacher likes you to leave the answers though.