An equation of the parabola is y=

Question

P(-1,0) Q(2,0) V(1,-2)

but the vertex isn’t completely in the center of the parabola. Not sure what to do I don’t know how to find it without the vertex

Answer 1

While V is usually reserved to represent the vertex, in this case it is not. It is just another point on the parabola.

This will be a system of 3 equations with 3 variables.

We parabolas are presented by Quadratic equations which have a standard form of

y = ax² + bx +c

Each of the given points are on the same parabola, which means they are solutions to the same quadratic equation. So we can get the three equations for our system by plugging in each point into that standard equation

Point P 0 = a(-1)² + b (-1) + c

Point Q 0 = a (2)² + b(2) + c

Point V -2 = a (1)² + b (1) + c

Simpliy...

EQ1 0 = a - b + c

EQ2 0 = 4a + 2b + c

EQ3 -2 = a + b + c

By elimination we can subtract EQ2 from EQ1 to get

0 = -3a - 3b

Now we need to involve EQ 3 with EQ 1 or 2 and eliminate c again.

EQ 2 - EQ 3 =...

2 = 3a + b

We now have a system of equations with 2 variables, that you can solve for a and b

0 = -3a - 3b

2 = 3a + b

After finding a and b you can plug them into one of the original equations to solve for c.

Then rewrite y = ax² + bx + c with a,b and c plugged in.

Answer 2

You are making the assumption that point V is the vertex. I doubt it. The only way to do something like that is to rotate the parabola like this:

And that is not typically taught in an Algebra 2 class so I suspect that something is in error, someone made a mistake, or point V is on the parabola but not the vertex.

Answer 3

The vertex is not needed.

0 = a(-1 - h)² + k

0 = a(1 - h)² + k

-2 = a(1 - h) + k

Solve the system for a, h, and k