Question

What is the vertex of `y=3(x-2)^2+1`?

Answer 1

`"vertex" -> (x,y) -> (2,1)`

Explanation:

`color(brown)("Introduction to idea of method.")`

When the equation is in the form `a(x-b)^2+c` then `x_("vertex")=(-1)xx(-b)`

If the equation form had been `a(x+b)^2+c` then `x_("vertex")=(-1)xx(+b)`
`color(brown)(underline(color(white)(" ."))`

`color(blue)("To find "x_("vertex"))`

So for `y=3(x-2)^2+1 :`

`color(blue)(x_("vertex")=(-1)xx(-2)=+2)`
`color(brown)(underline(color(white)(" ."))`

`color(blue)("To find "y_("vertex"))`
Substitute +2 into the original equation to find `y_("vertex")`

So `y_("vertex")=3((2)-2)^2+1`

`color(blue)(y_("vertex")=0^2+1 = 1)`

`color(brown)("Also notice this value is the same as the constant of +1 that is in the"` `color(brown)("vertex form equation.")`
`color(brown)(underline(color(white)(" ."))`

Thus: `color(green)("vertex" -> (x,y) -> (2,1))`

`color(purple)("~~~~~~~~~~~~~~~~~~~ Foot note ~~~~~~~~~~~~~~")`

Suppose the equation had been presented in the form of:

`y=3x^2-12x+13`

write as `y= 3(x^2-4x) +13`

If we carry out the mathematical process of
`(-1/2) xx(-4)=+2 =x_("vertex")`

The -4 comes from the `-4x" in "(x^2-4x)`

`color(purple)(" ~~~~~~~~~~~~~~~~~~End Foot note~~~~~~~~~~~~~~~~~~~~~~~~")`