Question

Find A, B, and C in the quadratic function (y=ax^2+bx+c) with the points (0,0) (1, 130) (2, 240) (11, ?) (If possible explain how to get it also)

Answer 1

First, it passes through (0,0), right?

Answer 2

So, 0=a(0)^2+b(0)+c=c
Therefore, c=0

Answer 3

yeah but I need tp figure out what A and B are also

Answer 4

I know. We'll get there, soon enough

Answer 5

Now you have coordinates for three points: (0, 0), (1, 130), and 2, 240).
Plug them into the quadratic equation ax² + bx + c = y to get a system of three equations:
0²a + 0b + c = 0
1²a + 1b + c = 130
2²a + 2b + c = 240

Solve the system to get
a = -10
b = 140
c = 0
-10x² + 140x = y equation of the parabola.

A1 11 ,
y = -10·11² + 140·11
= 330

Answer 6

Well, that's your answer.

Answer 7

I mainly can't figure out how to solve the system :/ I have like 3 more questions like this, just with different numbers