How to Write Quadratic Equations in Vertex Form and Standard Form

Vertex form is $y=a(x-h{)}^2+k$, where $(h,k)$ is the vertex.

Here $h=1$ and $k=2$, so the equation starts as $y=a(x-1{)}^2+2$. Note the sign changes only inside the bracket: a vertex at $x=1$ gives $(x-1)$, while the y-value 2 is written exactly as it is.

Since $a=1$, the equation is $y=(x-1{)}^2+2$.

Vertex form is $y=a(x-h{)}^2+k$, where $(h,k)$ is the vertex.

Here $h=2$ and $k=-3$, so the equation starts as $y=a(x-2{)}^2-3$. Only the x-value changes sign inside the bracket; the y-value keeps its minus sign as the constant.

Since $a=2$, the equation is $y=2(x-2{)}^2-3$. The 2 multiplies the bracket, so it cannot be left out.

Start from the vertex (1, 2), so the equation is $y=a(x-1{)}^2+2$.

Substitute the point (2, 3): $3=a(2-1{)}^2+2$, which gives $3=a+2$.

Subtract 2 from both sides: $a=3-2=1$.

So the equation is $y=(x-1{)}^2+2$.

Start from the vertex (2, 3), so the equation is $y=a(x-2{)}^2+3$.

Substitute the point (3, 1): $1=a(3-2{)}^2+3$, which gives $1=a+3$.

Subtract 3 from both sides: $a=1-3=-2$.

So the equation is $y=-2(x-2{)}^2+3$.

Start from the vertex (1, −3), so the equation is $y=a(x-1{)}^2-3$.

Substitute the point (2, 0): $0=a(2-1{)}^2-3$, which gives $0=a-3$.

Add 3 to both sides: $a=3$.

So the equation is $y=3(x-1{)}^2-3$, keeping the vertex y-value $-3$ as the constant.

Write the equation as $y=a{x}^2+bx+c$ and substitute each point.

From (0, 1): $c=1$, because $x=0$ makes the other two terms zero.

From (1, 2): $a+b+1=2$, so $a+b=1$.

From (2, 5): $4a+2b+1=5$, so $4a+2b=4$, and dividing every term by 2 gives $2a+b=2$.

Subtracting $a+b=1$ from $2a+b=2$ gives $a=1$, and then $b=0$.

So the equation is $y=x^2+1$.