How do you graph #g(x) = x^2 - 4x + 2#?

1 Answer
Jun 6, 2018

See below.

Explanation:

Easiest way is to identify points of interest that help to give the graph some structure.

In general, a polynomial with power #2# is a quadratic, which appears as a parabola (U-shape) on a graph.

Some properties that are useful are intercepts, the axis of symmetry, and the concavity.

We can find the #y#-intercept by setting #x = 0#:

#g(0) = 2#

So we know that the graph intersects the #y#-axis at #(0,2)#.

We can determine an axis of symmetry that equally divides the parabola using the equation #x =-b/(2a)# for an equation of the form #ax^2+bx+c#.

#x = -(-4)/(2(1)) = 4/2 = 2#

So we know the parabola is symmetric about the line #x = 2#.

We can take this further and find the #y# associated with #x =2#, since this will be the minimum, or root of the parabola.

#g(2) = -2#

So we know that the graph has a minimum at #(2,-2)#.

We can determine whether the parabola faces upwards or downwards based on the sign of #a# for an equation of the form #ax^2+bx+c#.

In this case, it's positive. So our parabola has a minimum and opens upward.

This should be sufficient information to sketch the graph rather well. If you want more specificity, you can substitute generic values of #x# into the equation to find the points that lie on the curve.

graph{x^2 - 4x + 2 [-10, 10, -5, 5]}

You should notice that the graph is indeed symmetric about #x = 2# with a minimum of #(2,-2)# and passes through #(0,2)#.

It'd also be easy to deduce that it passes through #(4,2)# as well, since it is symmetric about #x =2# with the point #(0,2)#.