Question

Write a quadratic function that fits the given set of points.

(-2, 5), (0, -3), and (3, 0)

Answer 1

@CGGURUMANJUNATH can you help?

Answer 2

we'll need to make a system of equations to solve this:

Y = Ax^2 + Bx + C

given that you have 3 points, make three equations from the standard form of the quadratic equation written above.

I'll do the first point so you see what I mean:
(-2, 5) combined with Y = Ax^2 + Bx + C becomes
5 = A(-2^2) + B(-2) + C

you make up the next two equations for the following two points, and we'll go from there.

Answer 3

\({\bf (-2, 5), (0, -3), (3, 0)\qquad \qquad y = ax^2+bx+c}\\ \quad \\
\begin{array}{llll}
(-2, 5)\qquad x=-2\qquad y=5\\ \quad \\
5 = a(-2)^2+b(-2)+c\implies &\color{blue}{5=4a-2b+c}\\ \quad \\
(0, -3)\qquad x=0\qquad y=-3\\ \quad \\
-3 = a(0)^2+b(0)+c\implies -3=0+0+c\implies &\color{blue}{c=-3}\\ \quad \\
(3, 0)\qquad x=3\qquad y=0\\ \quad \\
0 = a(3)^2+b(3)+c\implies &\color{blue}{0=9a^2+3b+c}
\end{array}\)

Answer 4

-3 = 0^2 + 0 + C

0 = -3^2 + -3 + C

Answer 5

yes, just keep in mind that the last point is (3, 0), not (-3, 0)

Answer 6

oh okay
so 0 = 3^2 + 3 + C

Answer 7

notice the 2nd equation, pretty much gives us "c", so you can use that in the other 2 equations and make them a system of equations of 2 variables, then solve like you'd any of those

Answer 8

hmmm...actually... the 3rd one should be only 9a, not \(9a^2\) anyhow

Answer 9

\(\bf 4a-2b+c=5\implies 4a-2b+(-3)=5\implies 4a-2b=8\\ \quad \\
9a+3b+c=0\implies 9a+3b+(-3)=0\implies 9a+3b=3\)

Answer 10

So the answer is

0=9a+3b+c

Answer 11

yes \({\bf (-2, 5), (0, -3), (3, 0)\qquad \qquad y = ax^2+bx+c}\\ \quad \\
\begin{array}{llll}
(-2, 5)\qquad x=-2\qquad y=5\\ \quad \\
5 = a(-2)^2+b(-2)+c\implies &\color{blue}{5=4a-2b+c}\\ \quad \\
(0, -3)\qquad x=0\qquad y=-3\\ \quad \\
-3 = a(0)^2+b(0)+c\implies -3=0+0+c\implies &\color{blue}{c=-3}\\ \quad \\
(3, 0)\qquad x=3\qquad y=0\\ \quad \\
0 = a(3)^2+b(3)+c\implies &\color{blue}{0=9a+3b+c}
\end{array}\)

Answer 12

then you'd just need to find "a" and "b", once you find those 3 guys, a, b, c, then plug them in in the quadratic template of \(\bf y = ax^2+bx+c\)

Answer 13

Think you could help me with 5 more questions like this?

Answer 14

if I had the time, which I don't, I'd be dashing in 5mins :/

Answer 15

Yeah that's okay! thank you for all of your help!

Answer 16

yw