Question

How do you find the coordinates of the vertex `y= -3x^2 + 8x`?

Answer 1

`(4/3,16/3)`

Explanation:

The `color(blue)"standard quadratic function"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=ax^2+bx+c; a≠0)color(white)(2/2)|)))`

`"for "y=-3x^2+8x`

`a=-3,b=8" and "c=0`

The x-coordinate of the vertex can be found using.

`color(red)(bar(ul(|color(white)(2/2)color(black)(x_("vertex")=-b/(2a))color(white)(2/2)|)))`

`rArrx_("vertex")=-8/(-6)=4/3`

Substitute this value into the equation to obtain the corresponding y-coordinate.

`y_("vertex")=-3(4/3)^2+8(4/3)`

`color(white)(y_("vertex"))=(-3xx16/9)+32/3`

`color(white)(y_("vertex"))=-16/3+32/3`

`color(white)(y_("vertex"))=16/3`

`rArr"coordinates of vertex "=(4/3,16/3)`
graph{-3x^2+8x [-20, 20, -10, 10]}