# Vertical Shifts of Quadratic Functions

## Key Questions

• In order to find the y-intercept $b$ of any function $f \left(x\right)$ is $f \left(0\right)$.

So, the y-intercept of $f \left(x\right) = a {x}^{2} + b x + c$ is

$f \left(0\right) = a {\left(0\right)}^{2} + b \left(0\right) + c = c$.

The constant term c of a quadratic function is always its y-intercept.

I hope that this was helpful.

• I would start with its vertex, then move either to the right or to the left, then use a symmetry to draw the other half.

I hope that this was helpful.

Vertical shifts are indicated by a constant added to the base function ${x}^{2}$, this changes the y-coordinate of the vertex.
$f \left(x\right) = {x}^{2} + 2$ moves the function (vertex) up 2 units
$f \left(x\right) = {x}^{2} - 3$ moves the function (vertex) down 3 units