With a given parallel line and a given point on the line
we can use the point-line method: y-y0=m(x-x0)
where
y=mx+k is the given line, and
(x0,y0) is the given point.
Here
m=-10, k=-5, (x0,y0)=(-3,5)
=> the required line L is given by:
L: y-5=-10(x-(-3))
on simplification
L: y=-10x-30+5
L: y=-10x-25
Yes, it is possible. For example, let's say that point A is (0,4) and there's a reflection over the y-axis, then point A would still be (0,4).
Answer:
This function is an even-degree polynomial, so the ends go off in the same directions, just like every quadratic I've ever graphed. Since the leading coefficient of this even-degree polynomial is positive, the ends came in and left out the top of the picture, just like every positive quadratic you've ever graphed. All even-degree polynomials behave, on their ends, like quadratics.
Step-by-step explanation:
