Question

What is the equation, in standard form, of a parabola that contains the following points?

(–2, 18), (0, 2), (4, 42)


y = –2x^2 – 2x – 3
y = –3x^2 + 2x – 2
y = 3x^2 – 2x + 2
y = –2x^2 + 3x + 2

Answer 1

when x=0, we can narrow down the options to 2 of them

Answer 2

If you have multiple choice, it's easy enough to just test them all out.

Answer 3

if you wanna go the long route:

y = 2+ax+bx(x+2)
y = 2-8x+bx(x+2)
y = 2-8x+3x(x+2)

and simplify

Answer 4

Im confused . . .

Answer 5

Thanks! I was just thinking about that too... testing multiple choice answers is efficient in terms of getting through a hw or test, but if you don't have multiple choice, it's nice to know how to set it up.

Answer 6

if you are confused, youshould try to lets us know what it is that you are confused about.

Answer 7

I don't understand how to test them out ?

Answer 8

The points given are (x,y) coordinates that satisfy the equation.
For example, the point (-2,18) means that
18 = a(-2)^2 +b(-2) +c
when you put x=-2 and y=18 into the equation,
y=ax^2 +bx +c.

Answer 9

you are given 3 points; which correspond to x and y values: (x,y)

Answer 10

try equation 1, using (0,2)
\[y = –2x^2 – 2x – 3\]replace x with 0 and y with 2
\[2 = –2*0^2 – 2*0 – 3\]and see if the results are true
\[2 = – 3~:~false\]
therefore it is not equation 1

Answer 11

If you didn't have multiple choice options, you would have to put all three points into
ax^2 + bx +c and solve a system of three equations for a, b, and c.

Answer 12

... "have to" ? nah. but that would be a suitable method ;)

Answer 13

so when i use equation 2 it would be like this?
y = –3x^2 + 2x – 2
4=-3*42^-3*42-2 ?

Answer 14

you have the right idea, but it loks like you are using the point (4,42). so youve just replaced the wrong parts

Answer 15

things are simpler when x=0, so stick with that to narrow the options

Answer 16

So , I still dont understand how to figure out the answer ?

Answer 17

y = –3 x^2 + 2 x – 2 ; using (4,42) would be
42=-3*4^2 +2*4 - 2

Answer 18

but using the (0,2) is way simpler since anyting time 0 is 0

y = –3 x^2 + 2 x – 2 ; using (0,2) would be
2=-3*0^2 +2*0 - 2
2 = -2

since 2=-2 is false, it aint equation 2

Answer 19

Ohhh , I see . . Hold on .

Answer 20

y = 3x^2 – 2x + 2
2= 3*0^2-2*0+2
2= 2

Is this the answer , cause I think it is !

Answer 21

this is a "possible" answer

and so is the last equation.

so we would need to use one of the other points to verify which of them is correct

Answer 22

using (4, 42)

y = 3 x^2 – 2 x + 2
42 = 3*4^2 – 2*4 + 2
42 = 3*16 – 8 + 2
42 = 48 – 8 + 2 ; i do like this 3rd one, but lets try this point in
the last equation to be sure

Answer 23

using the same point as the last time: (4, 42)

y = –2 x^2 + 3 x + 2
42 = –2*4^2 + 3*4 + 2
42 = –2*16 + 12 + 2
42 = –32 + 12 + 2 ; yeah, that point wont work in equation 4; so equation 3 looks great

Answer 24

Thank you sooo much for helping me , I get it now ! (:

Answer 25

youre welcome, and good luck :)

Answer 26

Thanks (;

Answer 27

if that last point DID work in equation 4, then we would have had to use the only point left, and that one would have only fit equation 3

Answer 28

Yes , I see ! (=

Answer 29

Enter the points in your stat plot edit screen
then push stat, -> calc, quadreg and it gives you the equation.