What is the vertex form of the equation?
y=x^2-6x+8
I have the answer I just need to make sure I have the steps.
My answer: y=(x-3)^2+8-9=y
=(x-3)^2-1
looks good to me.
To convert the given quadratic equation, y = x^2 - 6x + 8, into vertex form, which is in the form y = a(x - h)^2 + k, follow these steps:
Step 1: Group the x-terms and the constant term.
In this case, we have y = (x^2 - 6x) + 8.
Step 2: Complete the square for the x-terms.
To complete the square, we need to take half of the coefficient of the x-term, square it, and add it to both sides of the equation. The coefficient of the x-term is -6, so we will take half of it, which is -3, and square it, giving us (-3)^2 = 9. Adding 9 to both sides:
y + 9 = (x^2 - 6x + 9) + 8.
Step 3: Rewrite the expression inside the parentheses as a perfect square trinomial.
The expression inside the parentheses, x^2 - 6x + 9, can be factored as (x - 3)^2.
y + 9 = (x - 3)^2 + 8.
Step 4: Simplify the equation.
To simplify, subtract 9 from both sides:
y = (x - 3)^2 + 8 - 9.
This gives us the final equation in vertex form:
y = (x - 3)^2 - 1.
So, the vertex form of the equation y = x^2 - 6x + 8 is y = (x - 3)^2 - 1.