Convert it to standard form by FOILing the expression:
(x+2)(3x+2)color(red)(\implies) 3x^2+2x+6x+4color(red)(\implies) 3x^2+8x+4
Now that’s it in the form ax^2+bx+c, find the x-coordinate of the vertex using -\frac{b}{2a}:
-\frac{b}{2a}color(red)(\implies)-\frac{8}{2\cdot 3}color(red)(\implies) -\frac{4}{3}
Plug that into the standard form equation to find the y-coordinate:
3x^2+8x+4 color(red)(\implies) 3(-\frac{4}{3})^2+8(-\frac{4}{3})+4color(red)(\implies)\frac{48}{9}-\frac{32}{3}+4 color(red)(\implies)\frac{48}{9}-\frac{96}{9}+\frac{36}{9}color(red)(\implies) -\frac{4}{3}
color(red)(\therefore) the vertex is (-\frac{4}{3},-\frac{4}{3})
To find the x-intercepts, set each parenthetical expression of the original form of the equation (in the question) to 0, and solve for x:
x+2=0\qquad,\qquad 3x+2=0
color(red)(\implies)x=-2\qquad,\qquad x=-\frac{2}{3}
color(red)(\therefore) the x-intercepts are (-2,0) and (-\frac{2}{3},0).
Finally, the y-intercept is (0,c)color(red)(\implies) (0,4)
Here’s the graph:
![Desmos]()
