Question

f(x)=3x^2-6x+1 find the axis of symmetry and the vertex then rewrite in vertex form

Answer 1

ANSWER

The axis of symmetry is x = 1

The vertex form is

$f(x)=3(x-1)^2-2$

EXPLANATION

The axis of symmetry is a vertical line through the vertex of the parabola. This vertical line has te value of the x-coordinate of the vertex, which is:

$\begin{gathered} \text{ For a function with the form} \\ f(x)=ax^2+bx+c \\ \text{ the x-coordinate of the vertex is:} \\ x_v=-\frac{2a}{b} \end{gathered}$

For this problem a = 3 and b = -6:

$x_v=-\frac{2\cdot3}{-6}=-\frac{6}{-6}=-(-1)=1$

To write the equation in vertex form we have to find also the y-coordinate of the vertex. This is the value of the function when we replace x by xv:

$f(1)=3\cdot1^2-6\cdot1+1=3-6+1=-3+1=-2$

Therefore the vertex is at point (1, -2).

The vertex form of a quadratic equation is:

$f(x)=a(x-x_v)^2+y_v$

So for this function, the vertex form is:

$f(x)=3(x-1)^2-2$