yay, calc
we need to take take the implicit derivitive
to evaluate the slope, we need the point
when the line crosses the x axis, the value of y is 0
find the x value when y=0 to find the point, (x,0)
0+0+4x=8
4x=8
x=2
(2,0) Is the point
so
the derivitive is
\(2xy+x^2 \frac{dy}{dx}+3y^2 \frac{dy}{dx}+4=0\)
minus 2xy both sides
\(x^2 \frac{dy}{dx}+3y^2 \frac{dy}{dx}+4=-2xy\)
minus 4 both sides
\(x^2 \frac{dy}{dx}+3y^2 \frac{dy}{dx}=-2xy-4\)
undistribute dy/dx
\(\frac{dy}{dx}(x^2+3y^2)=-2xy-4\)
divide both sides by x²+3y²
\(\frac{dy}{dx} = \frac{-2xy-4}{x^2+3y^2}\)
now find the slope at the point (2,0)
sub 2 for x and 0 for y
\(\frac{dy}{dx} = \frac{-2(2)(0)-4}{2^2+3(0)^2}\)
\(\frac{dy}{dx} = \frac{0-4}{4+0}\)
\(\frac{dy}{dx} = \frac{-4}{4}\)
\(\frac{dy}{dx} = -1\)
slope is -1
use point slope form
the equation of a line that has a slope of m and passes through the point (x1,y1) is y-y1=m(x-x1)
slope is -1 and point is (2,0)
y-0=-1(x-2)
y=-1(x-2)
or
y=-x+2
