Example

Graphing y=x2+6x9y = -x^2 + 6x - 9

To graph the parabola y=x2+6x9y = -x^2 + 6x - 9, apply the standard graphing procedure using the axis of symmetry, vertex, intercepts, and symmetry.

Identify the coefficients and direction. Here a=1a = -1, b=6b = 6, and c=9c = -9. Because aa is negative, the parabola opens downward.

Find the axis of symmetry. Use x=b2ax = -\frac{b}{2a}:

x=62(1)=62=3x = -\frac{6}{2(-1)} = -\frac{6}{-2} = 3

The axis of symmetry is the vertical line x=3x = 3.

Find the vertex. Substitute x=3x = 3 into the equation:

y=(3)2+6(3)9=9+189=0y = -(3)^2 + 6(3) - 9 = -9 + 18 - 9 = 0

The vertex is (3,0)(3, 0).

Find the y-intercept. Substitute x=0x = 0:

y=(0)2+6(0)9=9y = -(0)^2 + 6(0) - 9 = -9

The y-intercept is (0,9)(0, -9).

Find the symmetric point. The y-intercept (0,9)(0, -9) lies 3 units to the left of the axis of symmetry x=3x = 3. The point 3 units to the right of the axis is (6,9)(6, -9).

Find the x-intercepts. Set y=0y = 0 and solve:

0=x2+6x90 = -x^2 + 6x - 9

Factor out 1-1:

0=(x26x+9)0 = -(x^2 - 6x + 9)

Recognize the perfect square trinomial:

0=(x3)20 = -(x - 3)^2

Solve to get x=3x = 3. The only x-intercept is (3,0)(3, 0), which is the same as the vertex.

Connect the points (3,0)(3, 0), (0,9)(0, -9), and (6,9)(6, -9) with a smooth downward-opening curve.

The vertex and x-intercept coincide because the discriminant of 0=x2+6x90 = -x^2 + 6x - 9 equals zero (624(1)(9)=3636=06^2 - 4(-1)(-9) = 36 - 36 = 0), meaning there is exactly one solution and the parabola just touches the x-axis at the vertex.

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Updated 2026-04-21

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