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

1 Answer
Apr 27, 2018

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:
enter image source here
(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!