Vertex y=x^2-6x

Here's how to find the vertex of the parabola y = x^2 - 6x: **1. Completing the Square** * **Rearrange the terms:** y = x^2 - 6x * **Factor out a coefficient of 1 from the x terms:** y = 1(x^2 - 6x) * **Take half of the coefficient of the x-term, square it, and add it inside the parentheses.** Half of -6 is -3, and squaring it gives 9. Because we're adding it inside the parenthesis where it's multiplied by 1, we need to subtract 9 outside to keep the equation balanced. y = 1(x^2 - 6x + 9) - 9 * **Factor the perfect square trinomial:** y = 1(x - 3)^2 - 9 **2. Vertex Form** Now the equation is in vertex form: y = A(x - h)^2 + k, where the vertex is at the point (h, k). * In our case, A = 1, h = 3, and k = -9 **Therefore, the vertex of the parabola is (3, -9).** **Key Points** * The parabola opens upwards because the coefficient 'A' is positive. * The vertex is the lowest point on the graph.