Question

How do you graph the parabola `y=x^2-3x` using vertex, intercepts and additional points?

Answer 1

Vertex : `(1.5, -2.25)`, axis of symmetry: `x=1.5`
x intercepts :`x=0,x=3,y` intercept is `y=0`,
additional points: `(4,4), (-1,4) and (5,10) ,(-2,10)`

Explanation:

`y=x^2-3 x or y= x^2 -3 x +1.5^2 -1.5^2`

or `y= ( x -1.5)^2 -2.25`

This is vertex form of equation ,`y=a(x-h)^2+k ; (h,k)`

being vertex , here `h=1.5 ,k=-2.25,a=1`

Since `a` is positive, parabola opens upward.

Therefore vertex is at `(1.5, -2.25)`

Axis of symmetry is `x= h or x = 1.5`

x-intercepts are found by putting `y=0` in the equation

`y=x(x-3)or x(x-3)=0 :. x=0 and x=3`

x intercepts are `x=0,x=3`

y-intercept is found by putting `x=0` in the equation

`y=x^2-3 x or y=0 :. y` intercept is `y=0`

Additional points on parabola: `x=4 , y=4(4-3)=4` ,

`x=-1 , y=-1(-1-3)=4 :. (4,4) and (-1,4)`

`x=5 , y=5(5-3)=10` ,

`x=-2 , y=-2(-2-3)=10 :. (5,10) and (-2,10)`

graph{x^2-3x [-10, 10, -5, 5]} [Ans]