Recall that the equation for a line can be expressed in the form
y=mx+by=mx+b where m=(x_2-x_2)/(y_2-y_1)=("vertical rise")/("horizontal run")m=x2−x2y2−y1=vertical risehorizontal run
Your yy-intercept is the value of yy when x=0x=0. When you plug a zero into the above equation you get the following:
y=mx+by=mx+b
y=m(0)+by=m(0)+b, where we let x=0x=0
y=0+by=0+b, because anything times zero is zero
y=by=b
So the yy-intercept is the value of yy when you set x=0x=0. But you were given a problem where xx can never be 00. You were given that x=-3x=−3 and, clearly, -3 != 0−3≠0. Well, if x != 0x≠0 (because it's -3), then the yy-intercept does not exist.
What about the slope? Remember the slope is ("rise")/("run")riserun. If you want to know how "steep" a line is, you must divide your vertical distance by your horizontal distance. A typical line (not your question above) looks like this:
![enter image source here]()
In order to turn this into a vertical line, you would have to make the run part really short, and the rise part really big. The graph starts to look like this:
![enter image source here]()
So when you are given x=-3x=−3, you have a vertical line. A vertical line has zero run because there is simply no left or right steepness. Additionally, the rise becomes infinite because there is only up and down in a vertical line. It's all rise and no run! Thus, the slope, m=oom=∞.
