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

1 Answer
Aug 14, 2018

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]