Answer:
[tex]y=(x+1)^2-2[/tex]
Step-by-step explanation:
General polynomial formula of a quadratic equation
[tex]ax^2+bx+c[/tex]
Equation [tex]y= x^2+2x-1[/tex]
[tex]a= 1\\ b=2\\ c=-1[/tex]
The formula to find the x coordinate of the vertex is
[tex]V_{x}= \frac{-b}{2a}[/tex]
[tex]V_{x}=\frac{-2}{2(1)} \\ V_{x}= -1[/tex]
Now to find the y coordinate of the vertex we substitute the found value of the x coordinate in [tex]y= x^2+2x-1[/tex]
[tex]y= (-1)^2+2(-1)-1 \\ y= 1-2-1\\ y=-2[/tex]
The general vertex formula for any quadratic function is
[tex]y= a(x-x_{v})^2+y_{v}[/tex]
Replace
[tex]y= 1(x-(-1))^2+(-2)\\ y=(x+1)^2-2[/tex]
