Please write y=-x62-2x+3 in vertex form. Help would be appreciated thank you :)
y = -x^2 - 2x + 3
y = -(x^2 + 2x) + 3
now add 1 in the parens to make a perfect square
y = -(x^2 + 2x + 1) + 1 + 3
y = -(x+1)^2 + 4
y-4 = -(x+1)^2
For y-k = a(x-h)^2
the vertex is at (h,k)
so, your vertex is at (-1,4)
To write the equation y = -x^2 - 2x + 3 in vertex form, you need to complete the square. Follow these steps:
Step 1: Group the x^2 and x terms together.
y = (-x^2 - 2x) + 3
Step 2: Factor out the coefficient of x^2 from the first two terms.
y = -1(x^2 + 2x) + 3
Step 3: Complete the square inside the parentheses. Adding and subtracting the square of half the coefficient of x (in this case, 1) will allow you to complete the square.
y = -1(x^2 + 2x + 1 - 1) + 3
Step 4: Rewrite the equation, leaving space for the constant term that will balance out the expression inside the parentheses.
y = -1((x + 1)^2 - 1) + 3
Step 5: Expand the equation inside the parentheses.
y = -1(x + 1)^2 + 1 + 3
Simplify the equation further.
y = -1(x + 1)^2 + 4
So, the given equation y = -x^2 - 2x + 3 can be written in vertex form as y = -1(x + 1)^2 + 4. The vertex form of the equation provides information about the vertex, which in this case is (-1, 4).