Math · College algebra · Worked example

Find the vertex and axis of a parabola

For f(x) = 2x² − 8x + 3, find the axis of symmetry, the vertex, the minimum value, the vertex form and the intercepts.

Find the axis of symmetry

Here a = 2 and b = −8.

- \frac{- 8}{2 (2)} = 2

Substitute x = 2. Since a = 2 > 0, the parabola opens upward and −5 is the minimum value.

\begin{array}{l} f (x) = 2 x^{2} - 8 x + 3 \\ f (2) = - 5 \end{array}

With vertex (2, −5) and a = 2, f(x) = 2(x − 2)² − 5. Expanding it gives back the standard form.

\begin{array}{l} 2 {(x - 2)}^{2} - 5 \\ 2 x^{2} - 8 x + 3 \end{array}

The y-intercept is f(0) = 3. Setting the vertex form equal to 0 gives x = 2 ± √(5/2), about 0.42 and 3.58, placed symmetrically about x = 2.

\begin{array}{l} 2 {(x - 2)}^{2} - 5 = 0 \\ {(x - 2)}^{2} = \frac{5}{2} \\ x = 2 \pm \sqrt{\frac{5}{2}} \end{array}

Axis x = 2, vertex (2, −5) and minimum value −5; f(x) = 2(x − 2)² − 5, with y-intercept 3 and x-intercepts 2 ± √(5/2) ≈ 0.42 and 3.58.

Find the vertex of g(x) = −x² + 6x − 4, and say whether it is a maximum or a minimum.

Show the answer and explanation

x = −6/(2(−1)) = 3 and g(3) = −9 + 18 − 4 = 5. Since a = −1 < 0, the parabola opens downward.

\begin{array}{l} g (x) = - x^{2} + 6 x - 4 \\ g (3) = 5 \end{array}

Graph plots f with its vertex and intercepts marked.

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