Question

Given the function g(x) = x2 – 6x + 8, Identify the zeros of the function.State the coordinates of the vertex point.State the equation for the axis of symmetry.Express the function in vertex form.Compare g(x) to the function f(x) = x2.

Answer 1

Consider the given function,

`g(x)=x^2-6x+8`

Apply the quadratic formula to find the roots,

`\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=(6\pm2)/(2) \\ x=2,4 \end{gathered}`

Consider the following,

`\begin{gathered} g(x)=(x)^2-2(x)(3)+(3)^2-1 \\ g(x)=(x-3)^2-1 \end{gathered}`

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,

`\begin{gathered} y=\frac{\text{Coefficient of x }}{2* Coefficientofx^2} \\ y=(-6)/(2*1) \\ y=-3 \end{gathered}`

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.