Vertex form

f(x) = 2x² + 4x − 4 and f(x) = 2(x + 1)² − 6 are the same function, but only the second shows the vertex (−1, −6) at a glance.

Where is the lowest point of 2x² + 4x − 4?

Try this before reading on. Where is the lowest point of f(x) = 2x² + 4x − 4? A table gets there, slowly. f(0) = −4, f(−1) = −6 and f(−2) = −4. The two −4s sit at equal distances from x = −1, so the bottom is at x = −1 and the minimum is −6. That took three evaluations and a lucky guess. Written as 2(x + 1)² − 6, the same function shows (−1, −6) with no table at all.

Why the vertex of a(x − h)² + k is (h, k)

A quadratic function can also be written as f(x) = a(x − h)² + k. OpenStax calls this the standard form, and because the vertex appears in it directly it is also called the vertex form. The vertex is the point (h, k). The reason is short: (x − h)² is a square, so it is never negative, and it equals 0 exactly when x = h. If a > 0, the smallest possible value of a(x − h)² is 0, so the smallest value of f is k, reached at x = h. If a < 0 the same argument gives the largest value.

Watch the sign inside the brackets. In f(x) = (x − 3)² + 2 the vertex is (3, 2), with a positive 3. In f(x) = −3(x + 2)² + 4, which OpenStax uses as an example, x + 2 is the same as x − (−2), so h = −2 and the vertex is (−2, 4). Since a = −3 is negative, the parabola opens down, 4 is the maximum value, and the range is all y with y ≤ 4. The axis of symmetry is the vertical line x = h, here x = −2.

h = −b/(2a) first, then k = f(h)

When a function is given as ax² + bx + c, there is a quick way to find the vertex. The axis of symmetry is x = −b/(2a), so that is h. Then k is just the output at h: k = f(h). Back to the opening problem, f(x) = 2x² + 4x − 4. Here h = −4/(2 × 2) = −1, and k = f(−1) = 2(1) − 4 − 4 = −6. Keep the same a and write f(x) = 2(x + 1)² − 6. Expanding confirms it: 2(x² + 2x + 1) − 6 = 2x² + 4x + 2 − 6 = 2x² + 4x − 4.

Why −b/(2a)? Expand a(x − h)² + k and you get ax² − 2ahx + (ah² + k). Matching this with ax² + bx + c term by term gives b = −2ah, and solving for h gives h = −b/(2a). The same comparison shows that c = ah² + k. So the two forms carry the same information; one is convenient for reading the vertex and the other for reading the y-intercept c.

Another route is completing the square. Take x² − 6x + 7. Half of −6 is −3, and (−3)² = 9. Add and subtract 9: x² − 6x + 9 − 9 + 7 = (x − 3)² − 2. The vertex is (3, −2). Both methods must agree, and here they do: −b/(2a) = 6/2 = 3 and f(3) = 9 − 18 + 7 = −2.

Sliding y = x² until its vertex sits at (h, k)

The vertex form also tells you how to draw the graph from y = x². Take the parent curve, stretch it by a (and flip it if a is negative), move it h units sideways and k units up or down. The vertex (0, 0) of y = x² lands at (h, k). In the lab, drag the red vertex: the panel rewrites the vertex form and the general form at the same time, so you can watch b and c change while the shape stays the same.