How can I find the equation of a quadratic function with three pts?

Question

The points are:

(-6, 5)
(3, 26)
(0, 5)

Answer 1

finding a quadratic function with three pts its a bit more work.
these are the points
(-6, 5)
(3, 26)
(0, 5)
and the general quadratic function is y = ax^2 + bx + c
awhile a point is ( x, y )

so next step we plug it in
first point into the general quadratic function
5 = a (-6)^2 + b (-6) + c
5 = 36a - 6b + c

then repeat it for the other two and you get
26 = 9a + 3b + c
5 = c

now with all three equations, you have to solve for three variables, by substitution, subtraction, and etc.
5 = 36a - 6b + c
26 = 9a + 3b + c
5 = c

here c = 5, so
5 = 36a - 6b + 5
26 = 9a + 3b + 5

then
0 = 36a - 6b => 0 = 6a - b
21 = 9a +3b => 7 = 3a +b

.....7 = 3a +b
+...0 = 6a - b
.....7 = 9a
a = 7 / 9

plug back for b
0 = 6 (7 / 9) - b
b = 6 (7 / 9)
b = 14 / 3

now sub everything back in and you get the equation.
y = (7 / 9)x^2 + (14 / 3)x + 5

Answer 2

The idea is really simple. Basically take those three points and plug them into the standard form of an quadratic equation. This gives you a 3-variable, 3-equation system of equations for you to solve. So with your three points you have:

a(-6)^2 + b(-6) + c = 5
a(3)^2 + b(3) + c = 26
a(0)^2 + b(0) + c = 5

Simplify it and you get:

36a - 6b + c = 5
9a + 3b + c = 26
c = 5

Now just solve for a, b, and c. And voila you have your quadratic equation.

Answer 3

To begin with you know that you are adding 5.
If the equation is in standard form ax^2+bx+c=0
then c is 5.

I also notice that when x=-6 then y=5 so
ax^2+bx=0

Answer 4

y = ax^2 + bx + c
5 = 36a - 6b + c
26 = 9a + 3x + c
5 = c
36a - 6b = 0
6a - b = 0
9a + 3b = 21
3a + b = 7
9a = 7
a = 7/9
b = 42/9 = 14/3
y = (7/9)x^2 + (14/3)x + 5

7*9/9 + 3*14/3 + 5 = 7 + 14 + 5 = 26
7*36/9 - 6*14/3 + 5 = 28 - 28 + 5 = 5

Answer 5

tell me one thing a quadratic equation is that which have two solution how can i make a quadratic eqation for three points isn't all these answers are wrong

Answer 6

(x3+y3)/(2x2y)