Question

Solve the vertex and axis of symmetry of the equation. Show your work andexplain the steps you used to solve.f(x) = x2 - 6x + 4

Answer 1

The vertex is at (3, -4).

The axis of symmetry is at x=3.

Explanation:

Given the following equation:

`f(x)=x^2-6x+4`

We need to convert it to the vertex form of a quadratic function, which is represented by:

`\begin{gathered} f(x)=a(x-h)^2+k\text{ } \\ \text{where (h,k) is the vertex of the parabola} \\ x=h\text{ is the axi}s\text{ of simmetry} \end{gathered}`

Let's use completing the square method to convert f(x) into vertex form:

`\begin{gathered} \text{Standard form:} \\ ax^2+bx+c=0 \\ \text{Then, for:} \\ x^2-6x+4=0; \\ a=1,\text{ b=-6 and c=4} \end{gathered}`

As a first step, subtract the constant (c) from both sides of the equation:

`\begin{gathered} x^2-6x+4-4=0-4 \\ x^2-6x=-4 \end{gathered}`

Then, add the square of b/2 on both sides of the equation:

`\begin{gathered} x^2-6x+(-(6)/(2))^2=-5+(-(6)/(2))^2 \\ x^2-6x+9=-5+9 \\ x^2-6x+9=4 \\ (x-3)^2=4 \\ (x-3)^2-4=0 \end{gathered}`

Now, we get the function in vertex form:

`f(x)=(x-3)^2-4`

Hence, the vertex is at (3, -4).

Therefore, if the axis of symmetry is at x=h, substitute:

The Axis of symmetry is at x=3.