The requirement is that every element in the domain must be connected to one - and one only - element in the codomain.
A classic visualization consists of two sets, filled with dots. Each dot in the domain must be the start of an arrow, pointing to a dot in the codomain.
So, the two things can't can't happen is that you don't have any arrow starting from a point in the domain, i.e. the function is not defined for that element, or that multiple arrows start from the same points.
But as long as an arrow start from each element in the domain, you have a function. It may happen that two different arrow point to the same element in the codomain - that's ok, the relation is still a function, but it's not injective; or it can happen that some points in the codomain aren't pointed by any arrow - you still have a function, except it's not surjective.
Answer:
Slope: -3, apparent point: (-1, -2)
Step-by-step explanation:
Point-slope form means the equation of a line is given in the form of y - y1 = m(x - x1), where x1 and y1 are coordinates of a point that lies on the line and m is its slope. Keeping this in mind while looking at the given equation, you can see that -3 is where m goes, meaning the slope is -3. You can also see that 1 and 2 are in the places where - x1 and - y1 go (don't forget those minus signs!) so you take those two and make them an ordered pair for your apparent point of (-1, -2).
Answer:
cash payment journal
Step-by-step explanation:
y = - 3(x + 1 )² + 2
the equation of a quadratic in vertex form is
y = a(x - h)² + k
where (h, k) are the coordinates of the vertex and a is a multiplier
here vertex = (- 1, 2), hence
y = a(x + 1)² + 2
to find a substitute (1, - 10) into the equation
- 10 = 4a + 2 ⇒ 4a = - 12 ⇒ a = - 3
y = - 3(x + 1)² + 2 ← in vertex form