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

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Answer 1 · 2017-06-20T06:26:54

This is an upright parabola with vertex #(1, -4)#, #x#-intercepts #(-1, 0)#, #(3, 0)# and #y# intercept #(0, -3)#

Given:

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

Since the coefficient of #x^2# is positive, this will be an upright (U-shaped) parabola.

When #x=0# we find:

#y = (color(blue)(0)+1)(color(blue)(0)-3) = -3#

So the #y#-intercept is at #(0, -3)#

Notice that #y=0# when #x=-1# or #x=3#

So this function has #x#-intercepts #(-1, 0)# and #(3, 0)#

The parabola will be symmetrical about its axis, which will be a vertical line half way between these two #x#-intercepts, that is:

#x = 1/2((color(blue)(-1))+color(blue)(3)) = 1#

The vertex lies on the axis, so we can find its #y# coordinate by substituting the value #x=1# into the formula:

#y = (color(blue)(1)+1))(color(blue)(1)-3) = 2*(-2) = -4#

So the vertex is at #(1, -4)#

That's probably enough points and features to graph looking roughly like this (with axis and points indicated):

graph{(y-(x+1)(x-3))(x-1+0.00001y)((x+1)^2+y^2-0.01)(x^2+(y+3)^2-0.01)((x-1)^2+(y+4)^2-0.01)((x-3)^2+y^2-0.01) = 0 [-9.595, 10.405, -5.36, 4.64]}