Which is narrower?

Let us write these equations of parabolas in their vertex form i.e. #y=a(x-h)^2+k#, where #(h.k)# is the vertex and #a# is quadratic coefficient. The greater the quadratic coefficient, the narrower is the parabola.

#f(x)=2x^2+3x=2(x^2+3/2x)#

= #2(x^2+2xx3/4x+(3/4)^2)-2xx(3/4)^2#

= #2(x+3/4)^2-9/8#

and #g(x)=x^2+4=(x-0)^2+4#

To find whether a parabola is narrow or wide, we should look at the quadratic coefficient of the parabola, which is #2# in #f(x)# and #1# in #g(x)# and hence f(x)=2x^2+3x# is narrower

graph{(y-x^2-3x)(y-x^2-4)=0 [-21.08, 18.92, -6, 14]}

Let's graph them both and then see for sure. Here is #f(x)=2x^2+3x#:

graph{2x^2+3x [-10, 10, -5, 20]}

And this is #g(x)=x^2+4#

graph{x^2+4 [-10, 10, -5, 20]}

Why is it that #g(x)# is fatter than #f(x)#?

The answer lies in the coefficient for the #x^2# term. When the absolute value of the coefficient gets bigger, the graph gets narrower (positive and negative simply show the direction the parabola is pointing, with positive opening up and negative opening down).

Let's compare the graphs of #y=pmx^2, pm5x^2, pm1/3x^2#. This is #y=pmx^2#:

graph{(y-x^2)(y+x^2)=0 [-10, 10, -5, 5]}

This is #y=pm5x^2#

graph{(y-5x^2)(y+5x^2)=0 [-10, 10, -5, 5]}

And this is #y=pm1/3x^2#

graph{(y-1/3x^2)(y+1/3x^2)=0 [-10, 10, -5, 5]}