Question

How do you graph the parabola `y=(x-3)^2 +5` using vertex, intercepts?

Answer 1

Since the parabola is in the form of:

`y = a(x-h) + k`

The entire parabola translates 3 units to the right and 5 units up. Therefore `(3, 5)` is the vertex (minimum turning point). Because the minimum point is shifted up by 5, he graph does not 'touch' the `x`-axis. Therefore, no `x`-intercepts can be determined on a real plane.

However, we can determine the `y`-intercept by substituting in `x=0` into the original equation.

`y = (0-3)^2 + 5`

`y = 9 + 5`

`y = 14`

Therefore the y-intercept is `(0, 14)`

Plot the points `(0, 14)` and `(3, 5)` on the Cartesian plane, connect the points in a parabolic form, and you have your graph.

graph{y = (x-3)^2 + 5 [-14.24, 14.24, -7.12, 7.12]}