Question

Fan&medal to best help! What is the equation, in standard form, of a parabola that contains the following points?

(-2,18), (0,2), (4,42)

A. y=-2x^2-2x-3
B. y=-3x^2+2x-2
C. y=3x^2-2x+2
D. y=-2x^2+3x+2

Answer 1

By the points you've given the parabola would be opening down so your quadratic would definitely be negative.

Answer 2

It actually opens up so its positive, oops I'll check my graphing calculator

Answer 3

So definitely D

Answer 4

Using
$$
(-2,18), (0,2), (4,42)
$$
We can solve
$$
y=ax^2+bx+c
$$
for a, b and c.
Using (-2,18), x=-2 and y=18:
$$
18=a-2^2+b(-2)+c=4a-2b+c
$$
Using (0,2), x=0 and y=2:
$$
2=a0^2+b(0) + c=c
$$
So now we know, c=2.
Finally, for (4,42), x=4 and y=42:
$$
42=a4^2+b(4)+c=16a+4b+c
$$
Putting this all together:
$$
18=4a-2b+2\\
42=16a+4b+2
$$
Adding 2 times the 1st equation to the second we get:
$$
2(18)=2(4a-2b+2)\\
+\\
42=16a+4b+2\\
=\\
36+42=8a+16a-4b+4b+4+2\\
78=24a+6\\
24a=78-6=72\\
a=72/24=3
$$
So,
$$
18=4(3)-2b+2\\
2b=12-18+2=-4\\
b=-2
$$
We have found that a=3,b=-2 and c=2:
$$
y=3x^2-2x+2
$$


Make sense?