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

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Answer 1 · 2015-06-18T22:43:51

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

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

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

$y = a(x-x_0)^2 + y_0$ , where $a != 0$ is a constant and $(x_0, y_0)$ is the vertex.

The axis of symmetry is $x=x0$ .

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

$y = a(0-x_0)^2 + y_0 = ax_0^2 + y_0$

In our case $a=1$ , $x_0 = 3$ and $y_0 = 5$

So the vertex is $(x_0, y_0) = (3, 5)$ , 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

$y = ax_0^2+y_0 = 1*3^2+5 = 9+5 = 14$

that is $(0, 14)$ .

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

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