Question

Find the coordinates of the vertex and write the equation of the axis of symmetry. $$y=\frac{1}{6} x^{2}-\frac{1}{3} x+2$$

Short answer

The coordinates of the vertex are \((1, \frac{3}{2})\) and the equation of the axis of symmetry is \(x = 1\).

Step-by-step solution

Find x-coordinate of the Vertex

Use the formula \(x = -\frac{b}{2a}\) to find the x-coordinate of the vertex. Here \(a = \frac{1}{6}\) and \(b = -\frac{1}{3}\). Plug these values into the formula, which yields \(x = -\frac{-\frac{1}{3}}{2\times\frac{1}{6}} = 1\)

Find y-coordinate of the Vertex

Substitute the x-coordinate of the vertex (which is 1) into the given equation to find the y-coordinate. Substitute \(x = 1\) into \(y = \frac{1}{6}x^{2} - \frac{1}{3}x + 2\), which yields \(y = \frac{1}{6} - \frac{1}{3} + 2 = \frac{3}{2}\)

Write the coordinates of the Vertex

The coordinates of the vertex is given by \((x, y)\). Here \(x = 1\) and \(y = \frac{3}{2}\), so the vertex is at \((1, \frac{3}{2})\)

Find the equation of the axis of Symmetry

The axis of symmetry is given by \(x = -\frac{b}{2a}\), which is the same as the x-coordinate of the vertex. Hence, the equation of the axis of symmetry is \(x = 1\)