Alexandra draws a letter A which stands on the $x$-axis.
[list=a][*]The left side of the letter A lies along the line with equation $y=3x+6$. What is the $x$-intercept of the line with equation $y=3x+6$?
[*]The right side of the letter A lies along the line $L_2$ and the leter is symmetric about the $y$-axis. What is the equation of line $L_2$?
[*]Determine the are of the triangle formed by the $x$ axis and the left and right sides of the letter A.
[*]Alexandra completes the letter A by adding to Figure 1. She draws the horizontal part of the letter A along the line $y=c$, as in Figure 2. The area of the shaded region inside the letter A and above the line with equation $y=c$ is $\frac49$ of the total area of the region above the $x$ axis and between the left and right sides. Determine the value of $c$.[/list]
[b]Figure 1[/b]
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -4.8408113739622465, xmax = 5.491811096383217, ymin = -3.0244242161812847, ymax = 8.241467380517944;  /* image dimensions */
pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); 


Label laxis; laxis.p = fontsize(10); 
xaxis(xmin, xmax, EndArrow(6), above = true); 
yaxis(ymin, ymax, EndArrow(8.25),above = true); /* draws axes; NoZero hides '0' label */ 
 /* draw figures */
draw((0,6)--(-2,0), linewidth(2)); 
draw((-2,0)--(2,0), linewidth(2)); 
draw((2,0)--(0,6), linewidth(2)); 
label("$y=3x+6$",(-2.874280000573916,3.508459668295191),SE*labelscalefactor); 
label("$L_2$",(1.3754276283584919,3.5917872688624928),SE*labelscalefactor); 
label("$O$",(0,0),SW*labelscalefactor); 
 /* dots and labels */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]
[b]Figure 2[/b]
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -4.707487213054563, xmax = 5.6251352572909, ymin = -3.4577277391312538, ymax = 7.808163857567977;  /* image dimensions */
pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); 

draw((-1.114884596113444,2.6553462116596678)--(1.1148845961134441,2.6553462116596678)--(0,6)--cycle, linewidth(2)); 


Label laxis; laxis.p = fontsize(10); 
xaxis(xmin, xmax, EndArrow(6), above = true); 
yaxis(ymin, ymax, EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ 
 /* draw figures */
draw((0,6)--(-2,0), linewidth(2)); 
draw((-2,0)--(2,0), linewidth(2)); 
draw((2,0)--(0,6), linewidth(2)); 
label("$O$",(0,0),SW*labelscalefactor); 
draw((-1.114884596113444,2.6553462116596678)--(1.1148845961134441,2.6553462116596678), linewidth(2)); 
draw((-1.114884596113444,2.6553462116596678)--(1.1148845961134441,2.6553462116596678), linewidth(2)); 
draw((1.1148845961134441,2.6553462116596678)--(0,6), linewidth(2)); 
draw((0,6)--(-1.114884596113444,2.6553462116596678), linewidth(2)); 
fill((0,6)--(-1.114884596113444,2.6553462116596678)--(1.1148845961134441,2.6553462116596678)--cycle,black);
label("$y=c$",(1.4920862691527148,3.1251527056856054),SE*labelscalefactor); 
 /* dots and labels */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* yes i used geogebra fight me*/
[/asy]
1. **Finding the $x$-intercept of the line $y = 3x + 6$:**

   To find the $x$-intercept, we set $y = 0$ and solve for $x$:
   \[
   0 = 3x + 6 \implies 3x = -6 \implies x = -2
   \]

2. **Finding the equation of the line $L_2$:**

   Since the letter A is symmetric about the $y$-axis, the right side of the letter A will be a reflection of the left side. The left side is given by $y = 3x + 6$. Reflecting this line over the $y$-axis, we replace $x$ with $-x$:
   \[
   y = 3(-x) + 6 \implies y = -3x + 6
   \]

3. **Determining the area of the triangle formed by the $x$-axis and the left and right sides of the letter A:**

   The vertices of the triangle are at $(-2, 0)$, $(2, 0)$, and $(0, 6)$. The base of the triangle is the distance between $(-2, 0)$ and $(2, 0)$, which is $4$ units. The height of the triangle is the $y$-coordinate of the vertex $(0, 6)$, which is $6$ units. The area $A$ of the triangle is given by:
   \[
   A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 6 = 12
   \]

4. **Determining the value of $c$ such that the area of the shaded region inside the letter A and above the line $y = c$ is $\frac{4}{9}$ of the total area:**

   The total area of the triangle is $12$. The area of the shaded region is $\frac{4}{9}$ of the total area:
   \[
   \text{Shaded area} = \frac{4}{9} \times 12 = \frac{48}{9} = \frac{16}{3}
   \]

   Let $h_1$ be the height from the $x$-axis to the line $y = c$. The area of the smaller triangle with height $h_1$ and base $b_1$ is:
   \[
   \frac{1}{2} \times b_1 \times h_1 = \frac{16}{3}
   \]

   Since the triangles are similar, the ratio of the heights is the same as the ratio of the bases. Let $h = 6$ and $b = 4$:
   \[
   \frac{h_1}{h} = \sqrt{\frac{\text{Area of smaller triangle}}{\text{Area of larger triangle}}} = \sqrt{\frac{\frac{16}{3}}{12}} = \sqrt{\frac{16}{36}} = \frac{4}{6} = \frac{2}{3}
   \]

   Therefore, $h_1 = \frac{2}{3} \times 6 = 4$. The value of $c$ is the height from the $x$-axis to the line $y = c$:
   \[
   c = h - h_1 = 6 - 4 = 2
   \]

The final answer is $\boxed{2}$