Analyzing the Vertex of y=6x-x²: Maximum or Minimum?

To solve this problem, we need to determine whether the quadratic function $y = 6x - x^2$ has a minimum or maximum point.

Firstly, let's identify the general form of the quadratic equation, which is $y = ax^2 + bx + c$. For our function, we have:

• $a = -1$
• $b = 6$
• $c = 0$

The coefficient $a = -1$ is negative, indicating that the parabola opens downwards. A downward-opening parabola means that the function has a maximum point.

Next, we calculate the x-coordinate of the vertex, which gives the maximum point:

The formula for the x-coordinate of the vertex of a quadratic function $ax^2 + bx + c$ is:

$x = -\frac{b}{2a}$

Substituting the values of $a$ and $b$ into the vertex formula, we get:

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

Thus, the x-coordinate of the vertex is $x = 3$. We can substitute this back into the original equation to find the y-coordinate:

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

Therefore, the vertex of the parabola is at $(3, 9)$, and since the parabola opens downwards, this point represents the maximum point of the function.

We conclude that the quadratic function $y = 6x - x^2$ reaches a highest point.