Find expressions for the quadratic functions whose graphs are shown.?

Question

1. A parable that is greater then 0 (opening is up) starting on (3,0) passes through point (4,2). I got x^2-6x+9 is that correct.
2. A parable that le less than zero (opening is down) passes through (-2,2), (0,1), and (1, -2.5) I can not figure out how to find this one out? Some help.

IF 1 is not correct can you please explain how to get the it. Not sure if I am on the right track?

Answer 1

if u plug in x = 4, u will not get y = 2 but y = 1 in your expression, but u correctly got the vertex. just multiply your expression by 2.
or
y = a(x - 3)^2 + k
plug in the points that u have and solve for the unknown.
2-
since it passes thru (0, 1)
it has the form y = ax^2 + bx + 1.....(the constant term is 1)
Just plug in the other points and get 2 equations in 2 unknowns.
2 = 4a - 2b + 1
-2.5 = a + b + 1
Solve for a and b.

Answer 2

1. With a vertex at (3,0)
y - 0 = a(x - 3)^2
If y(4) = 2
a(4 - 3)^2 = 2
a = 2
y = 2(x - 3)^2

2. y = ax^2 + bx + c
4a - 2b + c = 2
0a + 0b + c = 1
a + b + c = - 2.5
4a - 2b = 1
2a + 2b = - 7
6a = - 6
a = - 1
b = - 2.5
y = - x^2 - 2.5x + 1
y = - (x^2 + (5/2)x - 1)
y = - (x^2 + (5/2)5x + (5/4)^2 - 25/16 - 1)
y = - (x^2 + 5/4)^2 + 25/16 + 1
y - 41/16 = - (x^2 + 5/4)^2

Answer 3

Really interesting question, looking forward to going through the answers

Answer 4

It depends