Question

How do you find the maximum or minimum of `y-x^2+6=9x`?

Answer 1

The minimum is `(-4.5, -26.25)`.

Explanation:

The maximum or minimum of a quadratic equation is the same as the vertex. First, we have to write the equation in standard form, or `y = ax^2 + bx + c`

To do this, let's make `y` by itself. Add `x^2` and subtract `6` from both sides of the equation:
`y - x^2 + 6 quadcolor(red)(+quadx^2 quad-quad6) = 9x quadcolor(red)(+quadx^2 quad-quad6)`

`y = x^2 + 9x - 6`

To find the `x` value of the vertex, we use the formula `x = -b/(2a)`.

Let's plug in the numbers:
`x = -9/(2(1))`

`x = -4.5`

To find the `y` value of the vertex, we substitute back in the value of `x` back into the equation:
`y = x^2 + 9x - 6`

`y = (-4.5)^2 + 9(-4.5) - 6`

`y = 20.25 - 40.5 - 6`

`y = -26.25`

Therefore, using the `x` and `y` values of the vertex, we know that the vertex is at `(-4.5, -26.25)`. To verify this, let's graph it:

(desmos.com)

Because in this situation the vertex is the smallest point of the graph, we know that is is a minimum.

As you can see, the vertex/minimum is indeed `(-4.5, -26.25)`.

Hope this helps!