Find the vertex/focus of the parabola

Question

X = y^2 - 6y + 3

Find the vertex/focus of the parabola

Answer 1

Because the equation is written with "y²" we can see that the graph of this equation will open left/right. The standard form of a parabola that opens left/right is

x = a(y-k)² + h, where (h,k) is the vertex, y = k is the axis of symmetry and the focus is (h+1/(4a),k).

We can find h and k by first completing the square of the given equation as follows:

x = y² - 6y + 3 original equation

x = (y² - 6y) + 3 group like terms

x = (y² - 6y + 9) + 3 - 9 because we are adding 9 to the square we need to subtract 9 at the end to

the same equation

x = (y - 3)² - 6 simplified

Here we can see that a = 1, h = -6 and k = 3.

The vertex is (h,k)---->(-6,3).

The focus is (h+1/4a,k)---->(-6+1/(4)(1),3)---->(-5 3/4,3).

Answer 2

Using my calc, I found the vert is at 3,-6.35.

X = y^2 - 6y + 3

=y(y-6)+3

x/y=y-3

h=0, k=3

Answer 3

Since the y, and not the x, is squared, we know this is a sideways parabola. If we complete the square on the given equation to switch it to vertex form, it becomes x = (y-3)² -6. The vertex is always (h, k), where the h value is the -6 and the k value is 3, so our vertex is positioned at ( -6 , 3). Since the "a" value in the equation ( i.e the coefficient in front of the squared part) is not written, we take it to be 1, so with this being a positive value for a, we know the parabola opens to the right, so the focus must be to the right of the vertex.

Using the formula p = 1/(4a) where p is the distance from vertex to focus, we can insert our "a" value of 1 into the formula and we get p = 1/(4·1) or p = 1/4. So this is how far we must move to the right from our vertex to get to our focus. If we add 1/4 to the x coordinate of our vertex, leaving the y-coordinate as is, we have our focus positioned at ( -5 3/4, 3 ).

Answer 4

Hello Alyssa!

To solve your algebra two problem we are going to put the quadratic polynomial into vertex or 'standard' form. It may seem tricky because this is written in terms of x as a function of y. Backwards of what we're typically used to.

Our quadratic is currently in general form: f(y) = ay^2 + b*y + c, a = 1, b = -6, c = 3

Let put it in vertex or standard form: f(y) = a(y-h)^2 + k, where h = -b/2a = 6/(2*1) = 3,

and k = f(h) = 3^2-6*3+3 = -6

So the vertex form is f(y) = (y-3)^2 - 6 . And per the definition of the vertex form, the vertex is located at y = h & x = -6 (backwards of what we're used to again).

Vertex is at y = 3 & x = -6

Hope this helps and reach out if you ever need more help!