SOLUTION: Find the minimum value of x^2-2x-8, if it has one.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Find the minimum value of x^2-2x-8, if it has one.      Log On


   



Question 992634: Find the minimum value of x^2-2x-8, if it has one.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Does the graph of:y+=+x%5E2-2x-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%5E2%2Bbx%2Bc), 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%5E2 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+=+%28-b%29%2F2a The a is the coefficient of the x%5E2 term and the b is the coefficient of the x term.

In your equation, a+=+1 and b+=+-2, so
x+=+-%28-2%29%2F2%281%29
x+=+2%2F2
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%5E2-2x-8 Substitute x+=+1.
y+=+%281%29%5E2-2%281%29-8
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%28400%2C400%2C-5%2C5%2C-10%2C5%2Cx%5E2-2x-8%29