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

1 Answer
Jun 30, 2015

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)