Question

help me please. using the axis of symmetry find the vertex for the follow quadratic function. f (x)=3x^2-6x+8

Answer 1

`P(1,5)`

Step-by-step explanation: Axis of symmetry is a vertical line that makes function symmetrical along either side:

In case of parabla function or:

`y(x)=3x^2-6x+8`

We get axial symmetry where the first derivate is zero, and in fact, that is the x value for vertex:

Therefore:

`\begin{gathered} f^(\prime)(x)=(3x^2-6x+8)^(\prime)=6x-6=0 \\ \therefore\rightarrow \\ x=(6)/(6)=1 \end{gathered}`

And the corresponding y-value is:

`f(1)=3(1)^2-6(1)+8=5`

Therefore vertex is at the point:

`P(1,5)`