Question

What is the vertex of `y= 6(x-3)^2-x^2-5x+3`?

Answer 1

`"vertex "=(41/10,-541/20)`

Explanation:

`"expand and express in standard form"`

`y=6(x^2-6x+9)-x^2-5x+3`

`color(white)(x)=6x^2-36x+54-x^2-5x+3`

`color(white)(y)=5x^2-41x+57larrcolor(blue)"in standard form"`

`"with "a=5,b=-41" and "c=57`

`"given the equation in standard form then the x-coordinate"`
`"of the vertex is"`

`•color(white)(x)x_(color(red)"vertex")=-b/(2a)`

`x_("vertex")=-(-41)/10=41/10`

`"substitute this value into the equation for y-coordinate"`

`y_("vertex")=5(41/10)^2-41(41/10)+57`

`color(white)(xxxx)=1681/20-3362/20+1140/20=-541/20`

`color(magenta)"vertex "=(41/10,-541/20)`