How do you graph #f(x)=x^2-2# and identify the x intercepts, vertex?

1 Answer
Aug 14, 2017

If you're familiar with parent functions, graph #f(x)=x^2# and move it down the y axis two units.
The solutions are #+-sqrt(2)#.

Explanation:

Since #f(x)=x^2# is the parent function for a parabola, #f(x)=x^2 -2# is just that parent moved down two units. So this is just a parabola with a vertex that is at #(0,-2)# instead of being at #(0,0)#.
If you're not familiar with parent functions, make a table. Plug in some #x# values and find what the associated #y# values are.

To find the #x# intercepts, first think about what that means. We also call the #x# intercepts "zeros" and "solutions". These occur where the function crosses the #x# axis. If a point is on the #x# axis, the #y# coordinate is zero.
Since #f(x)# is the same thing as the #y# value, plug #0# in for #f(x)#: #0=x^2-2#. Add 2 to both sides, and you get #2=x^2#. Take the square root of both sides, and #+-sqrt(2)=x#.