Question

How do you find the vertex and the intercepts for `y=2x^2-12x`?

Answer 1

`color(blue)( y_("intercept")" is at y=0")`

`color(blue)(x_("intercepts")-> x=0; x=4)`

`color(blue)("Vertex"->(x,y)=(3,-18))" "->" and is a minimum"`

Explanation:

Write as:`" "y=2x^2-12x +0`

`color(blue)("The y intercept is at y=0")`

y =0 at x=0

`color(brown)("so one of the x intercepts is at x=0")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
write as `y=2(x^2-6x)`

`color(blue)(x_("vertex") = (-1/2)xx(-6) = +3)`

`color(brown)(y=2x^2-12xcolor(green)(" "->" "y_("vertex")=2(3)^2-12(3) =-18))`

`color(blue)(y_("vertex")=-18)`

`color(blue)("Vertex"->(x,y)=(3,-18))`

`color(brown)("As the coefficient of "x^2" is positive then the general shape of")` `color(brown)("the graph is " uu)` `color(blue)(" Thus the vertex is a Minimum")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`x_("intercpt")->y=0`

`=> 0=2(x^2-6x)`

Divide both sides by 2

`=>0/2=2/2(x^2-6x)`

But `0/2=0" and "2/2=1`

`=>0=(x^2-6x)`

Factor out `x`

`=>0=x(x-4)`

For `y=0 ; x=0" and/or "x=+4`

`color(blue)(x_("intercepts")-> x=0; x=4)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~