Sketch quadratics from their form

A completed square gives the turning point

For y = (x-3)²-4, the square is smallest at x = 3. The minimum is −4 and the turning point (3,−4). The symmetry line is x = 3, and the graph opens upward. The y-intercept is found separately: at x = 0, y = 5.

For y = -(x + 2)² + 9, the maximum is 9 at x = -2 and the graph opens downward. The sign in the bracket is reversed when locating the input that makes the square zero.

A factored form gives the roots

For y = (x-1)(x-5), the roots are 1 and 5 and the symmetry line is halfway between them, x = 3. Substitute to find the turning point: y = −4. The y-intercept is 5. Thus this is the same curve as the first completed-square example.

For y = -(x-1)(x-5), the roots stay the same, but all y-values change sign: the vertex becomes (3,4) and the parabola opens downward. A sketch must show the actual y-intercept as well as the roots and symmetry.

Check your understanding 1 Read both coordinates

Give the turning point, opening direction, y-intercept and roots of y = -(x-2)² + 1.

Your answer or drawing on paper

Show answer

The maximum point is (2,1), opening downward. At x = 0, y = −3. For roots, (x-2)² = 1, so x = 1 or 3. The symmetry line is x = 2.

A root need not exist in the real graph

y = (x-2)² + 1 has minimum 1, so it never reaches the x-axis. Do not invent roots because many earlier examples had two. A completed square can establish that absence directly.