How do you graph #y = x^2 + 3x + 2#?

1 Answer
May 1, 2016

The obvious method, but the least interesting is to choose #x# values, work out the corresponding #y# values and then plot the points. At least 5 or 7 points would be required to ensure some accuracy.

Explanation:

However, a more interesting method is to determine specific important points on the graph using algebra and then plot only those. The points are: the #y -# intercept, the two # x -# intercepts and the turning point (TP), which lies on the axis of symmetry.

For the #y#-intercept make #x# = 0.
This will give the constant as the #y#-intercept. #y# = 2.

For the #x#-intercepts make #y# = 0 and factorise.
The #x#-intercepts are also known as the roots of the equation.

If #y = (x+2)(x+1)#, then #0 = (x+2)(x+1)#
So #x =-2, or x = -1#

The line of symmetry lies halfway between the #x-#intercepts, giving #((-2)+( -1))/2 = -1.5#
This can also be found from the formula #x = -b/2a#

The #y# value of the TP can be calculated by substituting #x= 1.5# into the original equation, giving #y = -1/4#.

Plot these 4 points (0,2) ; (-2,0) ; (-1, 0) and (-1#1/2, -1/4#).
Join them with a smooth hand-drawn line and you have the graph.