1Cademy - Graphing $$y = -x^2 + 6x

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Finding the Intercepts of a Parabola

Example

Graphing $y = -x^2 + 6x - 9$

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

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

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

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

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

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

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

The vertex is (3, 0).

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

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

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

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

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

$0 = -x^2 + 6x - 9$

Factor out $-1$ :

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

Recognize the perfect square trinomial:

$0 = -(x - 3)^2$

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

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

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

Updated 2026-04-21

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OpenStax Elementary Algebra

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