How do you graph #y = 1/2(x-3)(x+1)#?

#y=1/2(x-3)(x+1)#

Is a quadratic equation, so we know that the plot will be a parabola.

The equation as given is in factored form. If the factored form is written with all real numbers, it gives us the x-intercepts (roots) for the plot. If an equation is in factored form,

#y=a(x-r_1)(x-r_2)#

then, the x-intercepts for the graph are at #(0, r_1)#, and #(0, r_2)#. If #a# is positive, then the plot is concave up. If #a# is negative, then the plot is concave down.

Here, #a=1/2# is positive, so we must draw our parabola concave up. Also, #r_1=3# and #r_2=-1# so the x-intercepts for the plot are #(0, 3)#, and #(0,-1)#.

Finally, before we start to plot the parabola, it would be helpful to know the coordinates of the vertex of the parabola. Here's a trick! The #x#-coordinate of the vertex of a parabola is the arithmetic average of the x-intercepts (roots) for the parabola.

#x#-coordinate of the vertex = #(r_1+r_2)/2#

Here, the #x#-coordinate of the vertex = #(3-1)/2=1#. We can find the #y#-coordinate of the vertex by substituting this #x# value into our original equation.

#y# coordinate of vertex #=1/2(1-3)(1+1)=-2#

The coordinates for the vertex of this parabola are at (1, -2).

Now we are ready to plot this parabola.

graph{1/2(x-3)(x+1) [-5, 5, -5, 5]}