Question

PLEASE BE QUICK AND ACCURATE

Answer 1

1. Create a Table.

We know that after a dreary day of rain, the sun peeks through the clouds and a rainbow forms. So you notice the rainbow is the shape of a parabola as shown in the first figure. This parabola has the following equation:

$y=-x^2+36$

On the other hand, in the distance you notice that an airplane is taking off. As it ascends during take-off, it makes a slanted line that cuts through the rainbow at two points. So let's choose two points on the graph of the parabola to build up the equation of the line:

$\bullet \ If \ x=-5, then: \\ \\ y=-(-5)^2+36=11 \\ \\ So: \\ \\ P_{1}(-5,11) \\ \\ \\ \bullet \ If \ x=4, then: \\ \\ y=-(4)^2+36=20 \\ \\ So: \\ \\ P_{2}(4,20)$

Therefore, the equation of the line using these two points is:

$y-20=\frac{20-11}{4-(-5)}(x-4) \\ \\ y-20=\frac{9}{9}(x-4) \\ \\ y-20=x-4 \\ \\ y=x+20-4 \\ \\ y=x+16$

FOR THE PARABOLA, THE TABLE IS:

$Other \ two \ points: \\ \\ \bullet \ If \ x=0, then: \\ \\ y=-(0)^2+36=36 \\ \\ So: \\ \\ P_{3}(0,36) \\ \\ \\ \bullet \ If \ x=1, then: \\ \\ y=-(1)^2+36=35 \\ \\ So: \\ \\ P_{4}(1,35)$

So the table is:

$\left[\begin{array}{cc}x & y\\-5 & 11\\4 & 20\\0 & 36\\1 & 35\end{array}\right]$

FOR THE LINE, THE TABLE IS:

$Other \ two \ points: \\ \\ \bullet \ If \ x=0, then: \\ \\ y=x+16=0+16=16 \\ \\ So: \\ \\ P_{3}(0,16) \\ \\ \\ \bullet \ If \ x=1, then: \\ \\ y=x+16=1+17=17 \\ \\ So: \\ \\ P_{4}(1,17)$

So the table is:

$\left[\begin{array}{cc}x & y\\-5 & 11\\4 & 20\\0 & 16\\1 & 17\end{array}\right]$

2. Analyzing the functions:

2.1 Domain and Range:

• The Domain of both the quadratic function and the linear function is the set of all real numbers.  

• The range of the quadratic function is $(-\infty,36)$ while the range of the linear function is the set of all real numbers.

The domain represents the horizontal distance while the range represents the height. In this situation, all values don't make sense. First, the rainbow starts in a point on the earth's crust. Also, the airplane starts at a point on the runway, so this doesn't include all the real numbers. A similar thing happens to the height since the airplane turns into an horizontal way at a moment of its trip and this doesn't include all the real numbers.

2.2 x and y intercepts of the rainbow.

x-intercepts:

The x intercepts are the points at which the graph passes through the x-axis, that is, when $y=0$. So:

$-x^2 + 36=0 \\ \\ x^2=36 \\ \\ x=\pm\sqrt{36} \\ \\ x=\pm 6$

So the x intercepts are $x_{1}=6 \ and \ x_{2}=-6$

y-intercepts:

The y-intercept is the point at which the graph passes through the y-axis, that is, when $x=0$. So:

$y=-x^2+36 \\ \\ y=-(0)^2+36 \\ \\ y=36$

So the y-intercept is $b=36$

In conclusion, if we take the x-axis as the ground, the x-intercepts represent the beginning and the end of the rainbow while the y-intercept represent its maximum height.

2.3 Is the linear function positive or negative

The equation of this line is:

$y=x+16$

As you can see, the slope $m=1$ is positive, therefore the line I've created is positive. This means as x increases y increases, but how do x and y increase? Well, they increase as the airplane is taking off, so this means the horizontal distance and the height increase.

2.4 What are the solutions or solution to the system of equations created?

The system of equations we created is:

$\left\{ \begin{array}{c}y=x+16\\y=-x^{2}+36\end{array}\right.$

Remember that in Part 1, chose two points on the graph of the parabola to build up the equation of the line, so the solutions to this system is indeed those two points:

$P_{1}(-5,11) \ and \ P_{2}(4,20)$

They represents that the airplane cuts through the rainbow at those two points.

3. Create a piecewise function.

A piecesewise function is a function defined by two or more equations over a domain. Our piecewise function is shown in the second figure below and is defined by:

$f(x)=\left\{ \begin{array}{c}x+16\quad if\quad-16\le x\le-5\\-x^{2}+36\quad if\quad-5\le x\le4\\x+16\ \quad if\quad4\le x\le24\\40\quad if\quad x\ge24\end{array}\right.$

To graph this function, let's follows these steps:

STEP 1: Graph $y=x+16$ from -16 to -5. This represents the graph of the trajectory of the airplane before touching the rainbow.

STEP 2: Graph $y=-x^2+36$ from -5 to 4. This is represents the graph of the rainbow as the airplane passes under its path.

STEP 3: Graph $y=x+16$ from 4 to 24. This is represents the graph of the trajectory of the airplane after touching the second point of the rainbow.

STEP 4: Graph $y=x+16$ from 24 and forward. This is represents the graph of the trajectory of the airplane when turning into an horizontal way.