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

1 Answer
George C.
Aug 1, 2015

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]}

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