#1
The two functions are:
and
so the difference is that -2. Adding or subtracting a constant moves a function up (if the constant is positive) or down (if the constant is negative). So adding -2 means the function will shift down 2 units. That is, the last choice given.
#2
The horizontal distance the rocket travels is given by x. As the rocket is launched and travels a parabolic path up into the sky and back down it moves a certain distance (horizontally) away from where it started. The rocket stops traveling when it hits the ground (when the height y is equal to 0). Therefore, we are being asked to find x when y = 0. We are being asked to solve the equation:
. To solve this we can use the quadratic formula.
The quadratic formula is:
. We need to determine the values of a, b and c from the equation we are trying to solve. a is the coefficient (number in front of)
, b is the coefficient of x and c is the constant (the number by itself). So in this problem we have:
a=-.06
b=9.6
c=5.4
We need to plug these values into the formula and simplify. Since we are rounding to the nearest hundredth (two decimal places) I would carry out your calculations to at least 4 decimal places and then round at the end. This reduces errors due to rounding. Here is how we use the formula:
![x={ \frac{-9.6(plus minus) \sqrt {9.6^{2} -4(-.06)(5.4)}}{2(-.06)} }](https://tex.z-dn.net/?f=x%3D%7B%20%5Cfrac%7B-9.6%28plus%20minus%29%20%5Csqrt%20%7B9.6%5E%7B2%7D%20-4%28-.06%29%285.4%29%7D%7D%7B2%28-.06%29%7D%20%7D)
![x= \frac{-9.6plusminus \sqrt{92.16+1.296} }{-.12}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B-9.6plusminus%20%5Csqrt%7B92.16%2B1.296%7D%20%7D%7B-.12%7D%20)
![x= \frac{-9.6plusminus 9.6672}{-.12}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B-9.6plusminus%209.6672%7D%7B-.12%7D)
Before we continue we need to recognize that our answer, being a distance, has to be positive. So even though there are two solutions to the equation only the positive one is correct. We arrive at this answer as follows:
<span>(-9.6-9.6672643)/-.12=160.56
c)
</span>This part is done the same way as part b. Again, the horizontal distance the rocket travels is given by x. As the rocket is launched and travels a parabolic path up into the sky and back down it moves a certain distance (horizontally) away from where it started. The rocket stops traveling when it hits the ground (when the height y is equal to 0). Therefore, we are being asked to find x when y = 0. We are being asked to solve the equation:
.
To solve this we can use the quadratic formula.
The quadratic formula is:
. We need to determine the values of a, b and c from the equation
we are trying to solve. a is the coefficient (number in front of)
, b is the coefficient of x and c is the constant (the
number by itself). So in this problem we have:
a=-.02
b=.8
c=37
We need to plug these values into the formula and simplify. Since we are rounding to the nearest hundredth (two decimal places) I would carry out your calculations to at least 4 decimal places and then round at the end. This reduces errors due to rounding. Here is how we use the formula:
![x={ \frac{-.8(plus minus) \sqrt {.8^{2} -4(-.02)(37)}}{2(-.02)} }](https://tex.z-dn.net/?f=x%3D%7B%20%5Cfrac%7B-.8%28plus%20minus%29%20%5Csqrt%20%7B.8%5E%7B2%7D%20-4%28-.02%29%2837%29%7D%7D%7B2%28-.02%29%7D%20%7D)
![x= \frac{-.8plusminus \sqrt{.64+.2.96} }{-.04}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B-.8plusminus%20%5Csqrt%7B.64%2B.2.96%7D%20%7D%7B-.04%7D%20)
![x= \frac{-.8plusminus 1.8973}{-.04}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B-.8plusminus%201.8973%7D%7B-.04%7D)
Before we continue we need to recognize that our answer, being a distance, has to be positive. So even though there are two solutions to the equation only the positive one is correct. We arrive at this answer as follows:
<span>(-.8-1.8973)/-.04=67.43</span>
The two functions are:
#2
The horizontal distance the rocket travels is given by x. As the rocket is launched and travels a parabolic path up into the sky and back down it moves a certain distance (horizontally) away from where it started. The rocket stops traveling when it hits the ground (when the height y is equal to 0). Therefore, we are being asked to find x when y = 0. We are being asked to solve the equation:
The quadratic formula is:
a=-.06
b=9.6
c=5.4
We need to plug these values into the formula and simplify. Since we are rounding to the nearest hundredth (two decimal places) I would carry out your calculations to at least 4 decimal places and then round at the end. This reduces errors due to rounding. Here is how we use the formula:
Before we continue we need to recognize that our answer, being a distance, has to be positive. So even though there are two solutions to the equation only the positive one is correct. We arrive at this answer as follows:
<span>(-9.6-9.6672643)/-.12=160.56
c)
</span>This part is done the same way as part b. Again, the horizontal distance the rocket travels is given by x. As the rocket is launched and travels a parabolic path up into the sky and back down it moves a certain distance (horizontally) away from where it started. The rocket stops traveling when it hits the ground (when the height y is equal to 0). Therefore, we are being asked to find x when y = 0. We are being asked to solve the equation:
The quadratic formula is:
a=-.02
b=.8
c=37
We need to plug these values into the formula and simplify. Since we are rounding to the nearest hundredth (two decimal places) I would carry out your calculations to at least 4 decimal places and then round at the end. This reduces errors due to rounding. Here is how we use the formula:
Before we continue we need to recognize that our answer, being a distance, has to be positive. So even though there are two solutions to the equation only the positive one is correct. We arrive at this answer as follows:
<span>(-.8-1.8973)/-.04=67.43</span>
4
0