How would you find the vertex form of the parabola graph in the image attached?
The vertex form of a parabola is given by,
`y = j(x-h)^2+k`
But we can write the equation as a a general quadratic equation and then convert to vertex form.
`y = ax^2+bx + c`
We have to find, a, b and c.
1. at `x = 0, y = 5`
`5 = axx0^2+bxx0+c`
`c = 5`
2. at `x = -3, y = 2`
`2 = axx(-3)^2+bxx(-3)+5`
`2 = 9a-3b+5`
`3a -b = -1`
3. at `x = 3, y =2`
`2 = axx3^2+bxx3+5`
We can solve for a and b,
by adding, `6a = -2 and a=-1/3`
by substracting, `2b=0 and b =0`
Therefore the equation is,
`y = -1/3x^2+5`
This is also equivalent to the vertex form,
where `j = -1/3` ` , h = 0 and k = 5`
The vertex form is,
`y = -1/3(x-0)^2+5`
Vertex form of a parabola is
`y = j(x - h)^2 + k`
Where, the vertex is located at (h,k)
Our goal is to figure out what j, h, and k are from looking at the graph.
We can look at the graph and see that the vertex is located at (0,5). So that means that h = 0, and k = 5.
We can rewrite our equation with this new information.
`y = j(x-0)^2 + 5`
Now we just need to find j.
That dot on the right of the graph tells us that when x = 3, y = 2.
So we can plug x = 3, y = 2 into our equation, this will let us solve for the last variable in our equation, j.
`2 = j(3-0)^2 + 5`
`2 = j*9 + 5`
`-3 = j*9`
`j = -3/9 = -1/3`
Now we have found what j, h, and k. We can write our equation in vertex form.
`y = -1/3(x-0)^2 + 5`