Verified answer

You have QUADRATIC FUNCTIONS f(x) = (x+4) (3x+8) which graph as parabolas.

I assume you want each parabola's vertex, zero's(x-intercepts, if any), axis of symmetry and y-intercept(x=0).

You want to get the function in the format: f(x) = a(x-h)² + k where a≠0 (or you wouldn't have a

quadratic); (h, k) is the vertex, making x=h the axis of symmetry

f(x) = (x+4) (3x+8) =3x² +20x +32 = 3[x² +(20/3)x + { } ] + 32 -{ } = completing the sq

3[x² +(20/3)x + {20/6}² ] + 32 -{400/12} [where the braces add zero ]

=3 [x +(10/3)]² - (4/3)

For this parabola the vertex is (-10/3, -4/3); zeros when x=-4 or -8/3; x=-10/3 is axis of sym.;

the y-intercept is 32.

In a similar fashion for f(n) = (n-3) (6n-1)

The vertex is (19/12, -289/24); zeros when x=3 or 1/6; n=19/12 is axis of sym; the intercept when n is zero is 3