If: y = -8 -3x and y = -2 -4x -x^2
Then: -8 -3x = -2 -4x - x^2
Transposing terms: x^2 +x -6 = 0
Factorizing: (x-2)(x+3) = 0 => x = 2 or x = -3
Points of intersection by substitution are at: (2, -14) and (-3, 1)
Intersection
It works out that line 3x-y = 5 makes contact with the curve 2x^2 +y^2 = 129 at (52/11, 101/11) and (-2, -11)
Then the intersection is a hyperbola.
The intersection of a line and a plane can result in either a single point, if the line passes through the plane, or no intersection at all if the line is parallel to the plane and does not touch it. In some cases, if the line lies entirely within the plane, every point on the line will be an intersection point. Thus, the nature of the intersection depends on the relative positions of the line and the plane.
True
It passes through Quadrants II and IV. It also passes through the origin ... the point where the 'x' and 'y' axes cross. At that point, it's in all four quadrants.
A vertical line that passes through -2 on the x-axis.
y=13x
So this would be a verticle line which passes through the point at -3.
y=x
It is a parabola, which passes through the origin and is symmetric about the y axis.
y = -4x is a line with a slop of -4 and a y-intercept of 0 (it passes through the point (0,0).