Consider the given function,
[tex]g(x)=x^2-6x+8[/tex]Apply the quadratic formula to find the roots,
[tex]\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(1)(8)}}{2(1)} \\ x=\frac{6\pm\sqrt[]{36-32}}{2} \\ x=\frac{6\pm\sqrt[]{4}}{2} \\ x=\frac{6\pm2}{2} \\ x=2,4 \end{gathered}[/tex]Consider the following,
[tex]\begin{gathered} g(x)=(x)^2-2(x)(3)+(3)^2-1 \\ g(x)=(x-3)^2-1 \end{gathered}[/tex]This is the vertex form of the quadratic equation. And the vertex of the parabola is (3,1).
Consider that the axis of symmetry of a quadratic equation is calculated as,
[tex]\begin{gathered} y=\frac{\text{Coefficient of x }}{2\times Coefficientofx^2} \\ y=\frac{-6}{2\times1} \\ y=-3 \end{gathered}[/tex]Thus, the axis of symmetry is the vertical line y=-3.
There are various conclusions which you can make comparing the function g(x) and f(x). Note that the function f(x) has both roots equal to zero, while g(x) has roots 2 and 4.
So we can conclude that the function f(x) is tangent to the x-axis at the origin, while the function g(x) intersects the x-axis twice.
