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#.