Answer:
Vertex form: [tex]y=\frac{1}{2}(x-2)^2-3[/tex]
Standard Form: [tex]y=0.50x^2-2x-1[/tex]
Step-by-step explanation:
Well the vertex form of an equation is given in the form: [tex]y=a(x-h)^2+k[/tex] where (h, k) is the vertex, and by looking at the graph, you'll see the vertex is at (2, -3). So plugging this into the equation gives you: [tex]y=a(x-2)^2-3[/tex]. Now to find a which will determine the stretch/compression, you can substitute any point in (besides the vertex, because that'll result in (x-2) being 0). So I'll use the point (0, -1) which is the only point I think I can accurately determine by looking at the graph (besides (4, -1) since it's symmetric). Anyways I'll plug this in
Plug in (0, -1) as (x, y)
[tex]-1 = a(0-2)^2-3[/tex]
calculate inside the parenthesis
[tex]-1 = a(-2)^2-3[/tex]
square the -2
[tex]-1 = 4a-3[/tex]
Add 3 to both 3 to both sides
[tex]2 = 4a[/tex]
divide both sides by 4
[tex]a=\frac{1}{2}[/tex]
This gives you the equation: [tex]y=\frac{1}{2}(x-2)^2-3[/tex]
To convert this into standard form you simply expand the square binomial, you can use the foil method to achieve this, but it generally expands to: [tex](a+b)^2=a^2+2ab+b^2[/tex].
Original equation:
[tex]y=\frac{1}{2}(x-2)^2-3[/tex]
expand square binomial:
[tex]y=\frac{1}{2}(x^2-4x+4)^2-3[/tex]
Distribute the 1/2
[tex]y=0.50x^2-2x+2-3[/tex]
Combine like terms:
[tex]y=0.50x^2-2x-1[/tex]
