Question

How do you graph `f(x)=x^2`?

Answer 1

This is a vertical parabola - a sort of U shape - with vertex at `(0, 0)`, axis of symmetry `x=0`, passing through `(4, -2)`, `(1, -1)`, `(1, 1)` and `(2, 4)`.

Explanation:

Ultimately, to graph (almost) any function `f(x)` you can compute `f(x)` for several values of `x` to find some points `(x, f(x))` through which the graph passes.

In our case `f(-2) = (-2)^2 = 4` gives us `(-2, 4)`, `f(-1) = 1` gives us `(-1, 1)`, etc.

In the general case of quadratic functions of form `f(x) = ax^2+bx+c` you can reformulate to find the vertex, axis of symmetry and where the parabola intersects the axes.

For example,

`f(x) = ax^2+bx+c = a(x-(-b/(2a)))^2+(c-b^2/(4a))`

`=a(x-h)^2+k`

with `h = -b/(2a)` and `k = c-b^2/(4a)`

This is in vertex form: The vertex is at `(h, k)`.

In our particular example, `a=1`, `b = c = 0`, so these formulae simplify to give `(h, k) = (0, 0)`

graph{(y-x^2)((x+2)^2+(y-4)^2-0.02)((x+1)^2+(y-1)^2-0.02)(x^2+y^2-0.02)((x-1)^2+(y-1)^2-0.02)((x-2)^2+(y-4)^2-0.02)(y*0.00001+x) = 0 [-10.5, 9.5, -2.28, 7.72]}