Answer:
the vertex of a quadratic equation:
a*x^2 + b*x + c = y
Is located at:
x = -b/2a.
In this case, x = -2
Then:
-b/2a = -2.
And we also have that:
a*(-2)^2 + b*-2 + c = 3
Then we have two equations:
-b/2a = -2
a*(-2)^2 + b*-2 + c = 3
We have only 2 equations and 3 variables, so we can just give the variable c different values and we will find different quadratic equations that are not equivalent.
Let's begin with c = 0, our equations are:
-b/2a = -2
a*(-2)^2 + b*-2 + 0 = 3
In the first equation we can isolate b and get:
b = 4*a
Now we can replace this in the other equation and solve this for a.
a*4 + 4*a*-2 = 3
-4*a = 3
a = -3/4
then:
b = 4*a = 4*-3/4 = -3
The first quadratic equation is:
(-3/4)*x^2 - 3*x = y.
Now, to find the second let's use a different value of c.
c = 1
b = 4*a
a*(-2)^2 + b*-2 + 1 = 3
Let's solve in the same way as before:
a*4 + (4*a)*-2 + 1 = 3
-4*a = 2
a = 2/-4 = -1/2
b = 4*-1/2 = -2.
Then our equation is:
(-1/2)*x^2 - 2*x + 1 = y
For the last equation, let's use c = 2.
then our equations are:
b = 4*a
a*(-2)^2 + b*-2 + 2 = 3
Same as before, replace the first equation in the second.
a*4 + (4*a)*-2 + 2 = 3
-4*a + 2 = 3
-4*a = 1
a = -1/4
Then:
b = 4*(-1/4) = -1
Our third equation is:
(-1/4)*x^2 - 1*x + 2 = y
Now, below you can see all of them graphed.
Where red is the first one, blue the second and green the third one. (and as you can see, all of them share the vertex (-2,3) )
