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Find a quadratic equation from its vertex and y-intercept

Start with vertex form

If a parabola has vertex $(h, k)$ , write $f (x) = a (x - h)^{2} + k$ . Substitute a second known point to determine the remaining coefficient $a$ .

Problem

A parabola has vertex $(- 2, 5)$ and crosses the y-axis at $(0, 1)$ . Find its quadratic function.

f (x) = - (x + 2)^{2} + 5 = - x^{2} - 4 x + 1

Both forms describe the same graph.

With vertex $(- 2, 5)$ , vertex form is $f (x) = a (x + 2)^{2} + 5$ .

The point $(0, 1)$ lies on the graph, so $1 = 4 a + 5$ . Therefore $a = - 1$ .

The function is $f (x) = - (x + 2)^{2} + 5$ . Expanding gives $- x^{2} - 4 x + 1$ . At $x = 0$ it equals 1, as required.

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