What does the graph of f(x)=3x^2-2x+1 look like

Below.

Step-by-step explanation:

This is a parabola.

As the coefficient of x^2 is positive it will open upwards ( shaped like a 'U').

If we convert to vertex form we can find the line of symmetry  and the coordinates of the vertex:

f(x) = 3x^2 - 2x + 1

f(x) = 3(x^2 - 2/3 x) + 1

f(x) = 3[(x - 1/3)^2 - (1/3)^2] + 1

f(x) = 3(x - 1/3)^2 - 1/3 + 1

f(x) = 3(x - 1/3)^2 + 2/3

So the line of symmetry is  x = 1/3 and

the vertex ( the minimum of the curve) is at (1/3, 2/3).

maybe showing the graph would

step-by-step explanation:

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