A equation of the form #y=ax^2+bx+c# is a parabola, as such #y=x^2+3x+1# too is a parabola.

Another way is to look at general form of equation #Ax^2+Bxy+Cy^2+Dx+Ey+F=0#

If #B^2-4AC=0# and #A=0# or #C=0#, it is a **parabola**

#B^2-4AC<0# and #A=C#, it is a **circle**

#B^2-4AC<0# and #A!=C#, it is an **ellipse**

#B^2-4AC>0#, it is a **hyperbola**

Here #A=1#, #B=0# and #C=0# and hence #B^2-4AC=0# and #C=0#. Therefore, it is a parabola.

The easiest way to graph such an equation is to convert it into vertex form i.e. #y=a(x-h)^2+k#. In such a case #(h,k)# is the vertex (if #a<0# it is at maxima and if #a>o# it s at minima) and #x=h# is axis of symmetry of parabola.

#y=x^2+3x+1#

= #x^2+2xx3/2xx x+(3/2)^2-(3/2)^2+1#

= #(x+3/2)^2-5/4=(x-(-3/2))^2-5/4#

Hence, vertex is #(-3/2,-5/4)# and axis of symmetry is #x+3/2=0#.

We can then choose different values of #x# around #-3/2# and find #y# from #y=x^2+3x+1# to get points

#(0,1), (1,5),(2,11),(-1,-1),(-2,-1),(-3,1),(-4,5),(-5,11)#

and graph appears as follows:

graph{x^2+3x+1 [-12.745, 7.255, -1.88, 8.12]}