Question 992634


Does the graph of:{{{y = x^2-2x-8}}} have a maximumum and minimum?
Well, it may have a maximum or it may have a minimum, but it doesn't have both!
The graph of a quadratic equation ({{{y = ax^2+bx+c}}}), such as this, is called a "parabola" and a parabola has a maximum or a minimum depending upon whether it opens upwards (has a minimum) or opens downwards (has a maximum).
You can tell which way the parabola opens by inspecting the coefficient of the {{{x^2}}} term, which is +1 in your problem.
If this coefficient is positive, then the parabola opens upwards and the graph has a minimum.
If the coefficient is negative,then the parabola opens downwards and the graph has a maximum.
The maximum/minimum point (also known as the "vertex") can be found as follows:
The x-coordinate of this point is:
{{{x = (-b)/2a}}} The a is the coefficient of the {{{x^2}}} term and the b is the coefficient of the {{{x}}} term.
 
In your equation, {{{a = 1}}} and {{{b = -2}}}, so

{{{x = -(-2)/2(1)}}}

{{{x = 2/2}}}

{{{x = 1}}} This is the x-coordinate of the vertex.

To find the y-coordinate, substitute {{{x = 1}}} into the given quadratic equation and solve for {{{y}}}:

{{{y = x^2-2x-8}}} Substitute {{{x = 1}}}.

{{{y = (1)^2-2(1)-8}}}

{{{y = 1-2-8}}}

{{{y = -9}}}

The vertex (or the {{{minimum}}} in this case) is at ({{{1}}}, {{{-9}}})

Let's see what the graph looks like:

{{{graph(400,400,-5,5,-10,5,x^2-2x-8)}}}