What is the vertex of #y=4x-x^2 #?

1 Answer
Feb 13, 2018

The vertex is at the point #(2, 4)#.

Explanation:

If you write your quadratic in standard form, (#ax^2+bx+c#) then the #x#-coordinate of the vertex is calculated by #(-b)/(2a)#.

Let's convert that equation to standard form:

#y=4x-x^2#

#quad=-x^2+4x#

#quad=-1x^2+4x+0#

In this case, our #a# value is #-1#, the #b# value is 4, and the #c# value is #0#. This means the #x#-coordinate of the vertex is #(-4)/(2(-1))# which is #2#.

Now, to find the #y#-coordinate, simply plug #2# into the equation and see the value that it returns:

#y=-1x^2+4x+0#

#quad=>-1(2)^2+4(2)+0#

#quad=-1*4+8#

#quad=-4+8#

#quad=4#

Now we know that the vertex is at an #x# of #2# and a #y# of #4#, or the point #(2, 4)#.

You can check this by graphing the parabola: graph{4x-x^2 [-8, 12, -2, 8]}