Example9.2.4
Create a table of values for the quadratic function \(y=x^2-4x+1\) and use this table to create its graph.
We'll start by choosing \(x\)-values between \(-3\) and \(3\text{:}\)
\(x\) | \(y=x^2-4x+1\) | Point |
\(-3\) | \((-3)^2-4(-3)+1=22\) | \((-3,22)\) |
\(-2\) | \((-2)^2-4(-2)+1=13\) | \((-2,4)\) |
\(-1\) | \((-1)^2-4(-1)+1=6\) | \((-1,6)\) |
\(0\) | \(0^2-4(0)+1=1\) | \((0,1)\) |
\(1\) | \(1^2-4(1)+1=-2\) | \((0,1)\) |
\(2\) | \(2^2-4(2)+1=-3\) | \((2,-3)\) |
\(3\) | \(3^2-4(3)+1=-2\) | \((3,-2)\) |
This is a good start, but doesn't give us the whole picture. Based on the pattern we see in Table 9.2.5, we can see that the \(y\)-values will continue to increase. It will be helpful to continue our table past values of \(x=3\text{:}\)
\(x\) | \(y=x^2-4x+1\) | Point |
\(-3\) | \((-3)^2-4(-3)+1=22\) | \((-3,22)\) |
\(-2\) | \((-2)^2-4(-2)+1=13\) | \((-2,4)\) |
\(-1\) | \((-1)^2-4(-1)+1=6\) | \((-1,6)\) |
\(0\) | \(0^2-4(0)+1=1\) | \((0,1)\) |
\(1\) | \(1^2-4(1)+1=-2\) | \((1,-2)\) |
\(2\) | \(2^2-4(2)+1=-3\) | \((2,-3)\) |
\(3\) | \(3^2-4(3)+1=-2\) | \((3,-2)\) |
\(4\) | \(4^2-4(4)+1=1\) | \((4,1)\) |
\(5\) | \(5^2-4(5)+1=6\) | \((5,6)\) |