Plot a graph for the equation ` y=ax-bx^(2)` , where a and b are positive constants.

Explanation

To find the vertex,
`(dy)/(dx)=a-2bx=0 rArr x=(a)/(2b)`
`:. Y=a.(a)/(2b)-b((a)/(2b))^(2)=(a^(2))/(2b)-(a^(2))/(4b)=(a^(2))/(4b)`
`:. ` vertex is `((a)/(2b), (a^(2))/(4b))`.
Also when `x=0, y=0` so curve passes through origin.
[Aliter : Arrange the equation to make quadratic experssion a perfect square.
`(y)/(b)=-(x^(2)-(a)/(b)x)=-(x^(2)-(a)/(b)x+(a^(2))/(4b^(2))-(a^(2))/(4b^(2)))`
`=-(x^(2)-(a)/(b)x+(a^(2))/(4b^(2))) +(a^(2))/(4b^(2))`
`(1)/(b)(y-(a^(2))/(4b))=-(x-(a)/(2b))^(2)`
` :. (y-(a^(2))/(4b))=-b (x-(a)/(2b))^(2) " " ` ...(1)
Putting `x-(a)/(2b)=0` and `y-(a^(2))/(4b)=0`
we get vertex ` ((a)/(2b), (a^(2))/(4b))]`
So it is a parabola and its axis is `x=(a)/(2b)`.
So graph is,