How to Write Quadratic Equations in Vertex Form and Standard Form

The vertex is at $(1,2)$, so the equation starts as $y=a(x-1{)}^2+2$. Since $a=1$, the equation becomes $y=(x-1{)}^2+2$.

The vertex is at $(2,-3)$, so the equation starts as $y=a(x-2{)}^2-3$. Since $a=2$, the quadratic equation is $y=2(x-2{)}^2-3$.

Start with the vertex $(1,2)$, so the equation is $y=a(x-1{)}^2+2$. Substituting $x=2$ and $y=3$ gives $3=a(1{)}^2+2$, so $a=1$. Therefore, the equation is $y=(x-1{)}^2+2$.

Using the vertex $(2,3)$, the equation is $y=a(x-2{)}^2+3$. Substituting the point $(3,1)$ gives $1=a(1{)}^2+3$, so $a=-2$. Thus, the equation is $y=-2(x-2{)}^2+3$.

Start with the vertex $(1,-3)$, so the equation is $y=a(x-1{)}^2-3$. Substituting the point $(2,0)$ gives $0=a(1{)}^2-3$, so $a=3$. Therefore, the equation is $y=3(x-1{)}^2-3$.

Substitute the points into the general form $y=a{x}^2+bx+c$. From $(0,1)$, $c=1$. Using $(1,2)$ gives $a+b=1$. Using $(2,5)$ gives $2a+b=2$. Solving these equations gives $a=1$ and $b=0$. Thus, the equation is $y=x^2+1$.