Which equation has a graph that is a parabola with a vertex at (1,HI)?
Oy=(-12+1
O y--1-
O y=(«+ 1) +1
O y=x+1)-1

Which equation has a graph that is a parabola with a vertex at 1HI Oy121 O y1 O y 1 1 O yx11 class=

Respuesta :

Answer:

[tex]y = (x +1)^2 - 1[/tex]

Step-by-step explanation:

Given

[tex]Vertex = (-1,-1)[/tex]

Required

Determine the equation

The equation of a graph from a vertex is:

[tex]y = a(x - h)^2 + k[/tex]

Where

[tex]a = 1[/tex]

[tex](h,k) = (-1,-1)[/tex]

Substitute these values in [tex]y = a(x - h)^2 + k[/tex]

[tex]y = 1(x - (-1))^2 + (-1)[/tex]

Express -(-1) as 1

[tex]y = 1(x +1)^2 + (-1)[/tex]

Express +(-1) as -1

[tex]y = 1(x +1)^2 - 1[/tex]

[tex]y = (x +1)^2 - 1[/tex]

Hence, the equation is:

[tex]y = (x +1)^2 - 1[/tex]