What is the vertex of y=3(x-3)^2-x^2+12x - 15?

1 Answer
Jan 19, 2016

"Vertex "-> (x,y)->(3/2,15/2)

Explanation:

color(blue)("Method:")
First simplify the equation so that it is in standard form of:
color(white)("xxxxxxxxxxx)y=ax^2+bx+c

Change this into the form:
color(white)("xxxxxxxxxxx)y=a(x^2+b/ax)+c This is NOT vertex form

Apply -1/2xxb/a= x_("vertex")

Substitute x_("vertex") back into the standard form to determine
y_("vertex")
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Given:color(white)(.....) y=3(x-3)^2-x^2+12x-15

color(blue)("Step 1")

y=3(x^2-6x+9)-x^2+12x-15

y=3x^2-18x+27-x^2+12x-15

y=2x^2-6x+12 ...........................................(1)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Step 2")

Write as: y=2(x^2-3x)+12
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Step 3")

color(green)(x_("vertex") = (-1/2)xx(-3)=+3/2).........................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Step 4")

Substitute value at (2) into equation (1) giving:

y_("vertex")=2(3/2)^2-6(3/2)+12

y_("vertex")=18/4-18/2+12

y_("vertex")=18/4-36/4+12

color(green)(y_("vertex")=-9/2+12=15/2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
"Vertex "-> (x,y)->(3/2,15/2)->(1 1/2, 7 1/2)

Tony B