Question

What is the vertex of `y=-x^2-3x+9`?

Answer 1

Vertex: `(-1.5, 11.25)`

Explanation:

`y = -x^2 - 3x + 9`

To find the `x`-coordinate of the vertex of a standard quadratic equation (`y = ax^2 + bx + c`), we use the formula `(-b)/(2a)`.

We know that `a = -1` and `b = -3`, so let's plug them into the formula:
`x = (-(-3))/(2(-1)) = 3/-2 = -1.5`

To find the `y`-coordinate of the vertex, just plug in the `x`-coordinate back into the original equation:
`y = -(-1.5)^2 - 3(-1.5) + 9`

`y = -2.25 + 4.5 + 9`

`y = 11.25`

Therefore, the vertex is at `(-1.5, 11.25)`.

Here's a graph of this equation (desmos.com):

As you can see, the vertex is indeed at `(-1.5, 11.25)`.

Hope this helps!