How do you draw the parabola #y=3x(x+2)#?

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Answer 1 · 2017-12-17T16:27:09

graph{y=3x(x+2) [-9.5, 10.5, -3, 7]}

#y=3x(x+2)#

Distribute the #3x# into the parenthesis

#y=3x^2+6x#

For the quadratic #y = ax^2 + bx + c# the vertex #(h, k)# which is the same as #(x,y)# is found by solving for #h = –b/(2a)#

#y=3x^2+6x#

#a=3#
#b=6#
#c=0#

#h=-6/(2(3))=-6/6=-1#

#h=-1#

#(h,k) => (-1,k)#

Now, to find the value of #k# (your #y# value), plug the found #h# value #(-1) # into the original equation # y=3x^2+6x#

#y=3(-1)^2+6(-1)#
#y=3(1)-6#
#y=3-6#
#y=-3#

#(h,k)=>(-1,-3)#

#(-1,-3)# is the vertex of the equation

Answer 2 · 2017-12-17T16:41:17

Expansion on prior answers!

I wanted to give a quick derivation of why the vertex point is at #x = -b/(2a) # for anyone who understands differential calculus:

The vertext point is also know as the stationary point, where the functions gradient is #0#

Let #y = ax^2 + bx + c #

Using power rule:

# (dy)/(dx) = 2ax + b #

When gradient is 0:

#2ax + b = 0 #

#=> 2ax = -b #

#=> x = -b/(2a) #