Amissile is launched from the ground. its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds. after 1 second, the missile is 130 feet in the air; after 2 seconds, it is 240 feet in the air. find the height, in feet, of the missile after 11 seconds in the air.

h(11) = 330 ft

Step-by-step explanation:

We have 2 points with which to find the equation of the projectile.  The standard form for the quadratic, which is the function that models parabolic motion, is

The problem tells us that the missile was launched from the ground, so c (the initial height of the missile) is 0, so we can literally disregard the c in the standard form and use the 2 points to solve for a and b:

At the point (1, 130) we fill in the standard form to get:

which gives us **130 = a + b**.  Now on to the second point, (2, 240):

which gives us **240 = 4a + 2b**

Solve the first **equation for a:

a = 130 - b

and sub that into the second **equation:

240 = 4(130 - b) + 2b and simplify to

140 = b

Now that we know b, we can sub it in to solve for a:

a = 130 - b so a = 130 - 140 so a = -10.  Therefore, the equation of the projectile's motion is

(If you knew your Physics, you'd know that -10x^2 is replacing the pull of gravity in the metric system which is -9.8x^2; that's how we know that this is the correct equation.)

Subbing in 11 for x:

gives us that h(11) = 330 feet

the range of a function is the range of y-values for which it is defined, or exists. given that we are only looking at the y-values, we can already discount the first two answers as they deal with the domain of the function (range of x-values).

now, what we need to do is find the smallest and highest y-values for which the function is defined. if you'd like, you could take a ruler and start at the very bottom of the graph, find the lowest point, then move the ruler higher and higher until you reach the highest point. doing so either in your head or with an actual ruler, we can see that the lowest y-value is -5 and the highest y-value is -1.

the question now remains as to whether they are included in the range (fourth answer) or excluded (third answer). the key here is to recognise whether they are marked with an open or a closed circle. since both (-2, -5) and (2, -1) are marked with a closed circle, they are points that exist on the graph and are thus included in the range (note that if they were marked with an open circle like (4, -2) then that would signify that there is a break in the graph at that point).

thus, the fourth answer, (y | -5 ≤ y ≤ -1), is correct.

also note that usually it is good practice to scan your eye inbetween the highest and lowest point for any breaks in the graph - we do have one here at (4, -2), however -2 is still included in the range because there is another point that is included in the graph with the same y-value, ie. point (1, -2).

i hope that but if you require any further explanation don't hesitate to comment below.

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .