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

1 Answer

1 vote

Answer:


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)

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