Given:
#y=1/3(x+4)(x+1)#
Expand #(x+4)(x+1)#.
#y=1/3(x^2+5x+4)#
Distribute #1/3#.
#y=1/3x^2+5/3x+4/3# is a quadratic equation in standard form:
#y=ax^2+bx+c#,
where:
#a=1/3#, #b=5/3#, and #c=4/3#
The vertex is the maximum or minimum point of the parabola. The formula for the axis of symmetry gives us the x-coordinate of the vertex:
#x=(-b)/(2a)#
#x=(-5/3)/(2*1/3)#
#x=(-5/3)/(2/3)#
#x=-5/3xx3/2#
#x=-15/6#
#x=-5/2# or #2.5#
To find the y-coordinate of the vertex, substitute #-5/2# for #x# and solve for #y#.
#y=1/3(-5/2)^2+5/3(-5/2)+4/3#
#y=1/3(25/4)-25/6+4/3#
#y=25/12-25/6+4/3#
The least common denominator is #12#. Multiply #25/6xx2/2# and #4/3xx4/4# to get equivalent fractions. Since #n/n=1#, the numbers will change but the value of each fraction will not change.
#y=25/12-25/6xxcolor(red)2/color(red)2+4/3xxcolor(blue)4/color(blue)4#
#y=25/12-50/12+16/12#
#y=-9/12#
#y=-3/4# or #-0.75#
The vertex is #(-5/2,-3/4)# or #(-2.5,-0.75)#. Plot this point.
The y-intercept is the value of #y# when #x=0#. Substitute #0# for #x# and solve for #y#.
#y=1/3(0)^2+5/3(0)+4/3#
#y=4/3# or #~~1.333#
The y-intercept is #(0,4/3)# or #(0,~~1.333)#. Plot this point.
The x-intercepts are the values for #x# when #y=0#. Substitute #0# for #y# and solve for #x#.
#0=1/3x^2+5/3x+4/3#
Switch sides.
#1/3x^2+5/3x+4/3=0#
Multiply both sides by #3#.
#x^2+5x+4=0#
Factor #x^2+5x+4#.
#(x+1)(x+4)=0#
Set each binomial to zero and solve.
#x+1=0#
#x=-1#
#x+4=0#
#x=-4#
The x-intercepts are #(-1,0), (-4,0)#. Plot these points.
Additional point: #x=-5#
#x=-5# is the mirror of the x-coordinate of the y-intercept.
Substitute #-5# for #x# and solve for #y#.
#y=1/3(-5)^2+5/3(-5)+4/3#
#y=25/3-25/3+4/3#
Additional point: #(-5,4/3)# or #(-5,~~1.333)#. Plot this point.
Sketch a graph through the points. Do not connect the dots.
graph{y=(x^2)/3+(5x)/3+4/3 [-10, 10, -5, 5]}
