Answer:
The third choice is the one you want
Step-by-step explanation:
If we are to write the equation of a line perpendicular to WX, we first must determine what the slope of the WX is, because the line perpendicular to WX has a slope that is the flip of the slope of WX with the opposite sign. Solving for y takes care of finding the slope of WX:
2x + y = -5 so
y = -2x - 5
The slope is -2. That means that the reciprocal slope is 1/2. Using that slope along with the coordinates x = -1 and y = -2, we first write the line using point-slope form and then solve it for y. Start by filling in the m, the x value and the y value:

Getting rid of the double negatives gives us:

Distributing then gives us:

And finally solving for y (I am going to express the 2 on the left as 4/2 when I move it by subtraction in order to add those fractions):

And the final equation in slope-intercept form is:

The first equation
6x^2 -18x-24=0
-the graph opens up
6(x^2 -3x -4)=0
6(x^2 -3x +9/4-4-9/4)=0
6(x-3/2)-25x6/4=
6(x-3/2)-75/2
-Vertex is at (3/2, -75/2)
-Axis of symmetry is x=3/2
Second equation
3r^2 -16r -12=0
-the graph opens up
...
The solution to your problem is m>-2
Given that the number of years should be represented with x, the number of fish in the pond after x years should best be represented with f(x). The equation that would best show the given scenario in the problem above is,
f(x) = 500(2^x)
From the given, 500 is used as the initial population of the fish.