What is the vertex of # y= -(x-6)^2-3x^2-2x+3#?

1 Answer
Jun 19, 2017

( #1.25,-26.75#).

Explanation:

Your starting equation is:
#-(x-6)^2-3x^2-2x+3#

The easiest way to solve this is to expand the #(x-6)^2#, add everything up to get it into standard form, and then use the vertex equation for standard form to find the vertex.

Here is how you use the square method to multiply two binomials (A binomial is a thing with two terms; usually one variable and one definite number, like x-6.):

     x    -   6

x [ #x^2# | -6x]
-6[-6x | 36]

(apologies for bad formatting)

How you do this is basically you make a square, divide it into four smaller squares (Like the windows symbol), and put one binomial on top, and one on the left side vertically. Then, for each box, multiply the term of the binomial (The thing outside the box) on top of it and to the left of it.

#(x-6)^2# expanded is #x^2-12x+36#, which means that the full equation is #-(x^2-12x+36)-3x^2-2x+3#. That simplifies to:

#-x^2+12x-36-3x^2-2x+3#

Now, just add up the like terms.
#-x^2+(-3x^2) = -4x^2#
#12x+(-2x) = 10x#
#-36+3 = -33#

The whole equation in standard form ( #ax^2+bx+c# form) is #-4x^2+10x-33#.

The vertex equation, #(-b)/(2a)#, gives you the x-value of the vertex. Here, 10 is b and -4 is a, so we need to solve #(-10)/-8#. That simplifies to 5/4, or 1.25.

To find the y-value of the vertex, we need to plug the x-value into the equation.

#-4(1.25)^2+10(1.25)-33 = -4(1.5625)+12.5-33 = -6.25+12.5-33 = -26.75.#

The y-value of the vertex is -26.75, so the vertex is ( #1.25,-26.75#).

And to check it, here's the graph:
graph{y = -(x-6)^2-3x^2-2x+3 [0.061, 2.561, -27.6, -26.35]}