Question

How do you find the vertex of a parabola `y = -2(x+1)(x-5)`?

Answer 1

The vertex is at the point `(2,18)`, that is `x=2`, `y=18`

Explanation:

Consider a parabola represented by a quadratic polynomial
`y=ax^2+bx+c`
If `a>0` the parabola's horns are directed upwards and it reaches its minimum at some central point `x=x_0`.
If `a<0` the parabola's horns are directed downwards and it reaches its maximum at some central point `x=x_0`.
In both cases the X-coordinate of a vertex is `x_0` and its Y-coordinate is `ax_0^2+bx_0+c`.

Parabola is a symmetrical figure, its vertical axis of symmetry intersects the X-axis in exactly the same point where it reaches its minimum (for `a>0`) or maximum (for `a<0`) value, that is at point `x=x_0`.

If we are dealing with a quadratic polynomial that can be equal to zero at some values of `x`, that is if the equation
`ax^2+bx+c=0`
has solutions `x=x_1` and `x=x_2`,
the central point `x_0` lies exactly in the middle between these solutions `x_1` and `x_2`.

Therefore, in case we have two solutions mentioned above, all we have to do to find a vertex is
(a) find these two solutions `x_1` and `x_2` of an equation:
`ax^2+bx+c=0`
(b) find a midpoint between them:
`x_0=(x_1+x_2)/2`.

The given problem presents an easy case, when the solutions to an equation `-2(x+1)(x-5)=0` are, obviously, `x_1=-1` and `x_2=5`.
Hence, the midpoint between them is
`x_0=(-1+5)/2=2`

So, we have determined that X-coordinate of the parabola's axis of symmetry as `x_0=2`.
The Y-coordinate is `y_0_ = -2(2+1)(2-5)=18`.
So the vertex is at point `(2,18)`