# How to graph a parabola y=(x-3)^2 +5?

Jun 18, 2015

The equation $y = {\left(x - 3\right)}^{2} + 5$ is in vertex form, so you can quickly deduce that the vertex is at $\left(3 , 5\right)$ and axis of symmetry is $x = 3$

The parabola intercepts the $y$ axis at $\left(0 , 14\right)$.

#### Explanation:

Vertex form of the equation of a parabola with vertical axis is:

$y = a {\left(x - {x}_{0}\right)}^{2} + {y}_{0}$, where $a \ne 0$ is a constant and $\left({x}_{0} , {y}_{0}\right)$ is the vertex.

The axis of symmetry is $x = x 0$.

The intercept with the $y$ axis can be found by substituting $x = 0$ into the equation to get:

$y = a {\left(0 - {x}_{0}\right)}^{2} + {y}_{0} = a {x}_{0}^{2} + {y}_{0}$

In our case $a = 1$, ${x}_{0} = 3$ and ${y}_{0} = 5$

So the vertex is $\left({x}_{0} , {y}_{0}\right) = \left(3 , 5\right)$, the axis of symmetry is $x = {x}_{0} = 3$.

The parabola is U-shaped since $a > 0$. graph{(x-3)^2+5 [-18.58, 21.42, -1.76, 18.24]}

The intercept with the $y$ axis is where

$x = 0$ and

$y = a {x}_{0}^{2} + {y}_{0} = 1 \cdot {3}^{2} + 5 = 9 + 5 = 14$

that is $\left(0 , 14\right)$.

The parabola does not intercept the $x$ axis, since the minimum value of $y$ is $5$.