2. How is the graph of y = –2x² – 5 different from the graph of y = –2x²? (1 point)
It is shifted 5 units up.
It is shifted 5 units down.
It is shifted 5 units left.
It is shifted 5 units right.
3. A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation
y = –0.04x
2
+ 8.3x + 4.3 , where x is the horizontal distance, in meters, from the starting point on the roof and y is the
height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land?
Round your answer to the nearest hundredth meter. (1 point)
208.02 m
416.03 m
0.52 m
208.19 m
4. A physics student stands at the top of a hill that has an elevation of 56 meters. He throws a rock and it goes up
into the air and then falls back past him and lands on the ground below. The path of the rock can be modeled by the
equation y = –0.04x² + 1.3x + 56 where x is the horizontal distance, in meters, from the starting point on the top of the
hill and y is the height, in meters, of the rock above the ground.
How far horizontally from its starting point will the rock land? Round your answer to the nearest hundredth. (1 point)
56.00 m
24.54 m
57.04 m
57.26 m
How many real-number solutions does the equation have?
5. –7x² + 6x + 3 = 0 (1 point)
one solution
two solutions
no solutions
infinitely many solutions
6. How many real number solutions does the equation have?
y = 3x² + 18x + 27 (1 point)
one solution
two solutions
no solutions
infinitely many solutions
9. Find the solutions to the system.
y = x² – 5x + 2
y = –6x + 4 (1 point)
(2, 8) and (−1, 2)
(−2, 8) and (1, −2)
(−2, 16) and (1, −2)
no solutions
10. Find the solutions to the system.
y = x² + 8x + 2
y = 7x + 4 (1 point)
(−2, −10) and (1, 11)
(2, 10) and (−1, −11)
(−2, −34) and (1, 11)
no solutions
11. If an object is dropped from a height of 200 feet, the function h(t) = –16t² + 200 gives the height of the object
after t seconds. Approximately, when will the object hit the ground? (1 point)
200.00 seconds
184.00 seconds
3.54 seconds
0.78 seconds
12. A ball is thrown into the air with an upward velocity of 32 feet per second. Its height, h, in feet after t seconds is
given by the function h(t) = –16t² + 32t + 6. How long does it take the ball to reach its maximum height? What is the
ball’s maximum height? Round to the nearest hundredth, if necessary. (1 point)
Reaches a maximum height of 22 feet after 1.00 second.
Reaches a maximum height of 22 feet after 2.00 seconds.
Reaches a maximum height of 44 feet after 2.17 seconds.
Reaches a maximum height of 11 feet after 2.17 seconds.
13. A catapult launches a boulder with an upward velocity of 148 ft/s. The height of the boulder (h) in feet after t
seconds is given by the function h = –16t² + 148t + 30. How long does it take the boulder to reach its maximum
height? What is the boulder’s maximum height? Round to the nearest hundredth, if necessary. (1 point)
Reaches a maximum height of 30 feet after 9.25 second.
Reaches a maximum height of 640.5 feet after 4.63 seconds.
Reaches a maximum height of 1,056.75 feet after 4.63 seconds.
Reaches a maximum height of 372.25 feet after 4.63 seconds
2. The graph of y = –2x² – 5 is shifted 5 units down from the graph of y = –2x².
3. The rocket will land 208.02 meters horizontally from its starting point.
4. The rock will land 57.04 meters horizontally from its starting point.
5. The equation –7x² + 6x + 3 = 0 has two real-number solutions.
6. The equation y = 3x² + 18x + 27 has one real-number solution.
9. The solutions to the system y = x² – 5x + 2 and y = –6x + 4 are (2, 8) and (−1, 2).
10. The system y = x² + 8x + 2 and y = 7x + 4 has no solutions.
11. The object will hit the ground approximately at 3.54 seconds.
12. The ball reaches its maximum height of 22 feet after 1.00 second.
13. The boulder reaches its maximum height of 640.5 feet after 4.63 seconds.
2. The graph of y = –2x² – 5 is different from the graph of y = –2x² because it is shifted 5 units down. To understand this, note that the general format of a quadratic equation is y = ax² + bx + c. In this case, the equation y = –2x² has no value added or subtracted to it, so its graph is the standard parabola with vertex (0, 0). However, when 5 units are subtracted from the equation, as in y = –2x² – 5, the entire graph is shifted downward by 5 units.
3. To find how far horizontally the rocket will land in the given equation y = –0.04x² + 8.3x + 4.3, we need to determine the value of x when y = 0. This is because when the rocket lands, its height will be 0. So, we set y = 0 and solve the equation –0.04x² + 8.3x + 4.3 = 0. Using the quadratic formula or factoring, we find the solutions for x. The resulting x-values give us the horizontal distance for the rocket's landing point.
4. Similar to question 3, to find how far horizontally the rock will land in the given equation y = –0.04x² + 1.3x + 56, we need to find the x-values when y = 0. By setting y = 0 and solving the equation –0.04x² + 1.3x + 56 = 0, we can determine the horizontal distance at which the rock will land.
5. To determine the number of real-number solutions for the equation –7x² + 6x + 3 = 0, we can use the discriminant. The discriminant is given by the formula b² - 4ac for a quadratic equation in the form ax² + bx + c = 0. If the discriminant is greater than 0, the equation has two real solutions. If it is equal to 0, the equation has one real solution. If it is less than 0, the equation has no real solutions.
6. Similarly to question 5, to find the number of real number solutions for the equation y = 3x² + 18x + 27, we can use the discriminant.
9. To solve the system of equations y = x² – 5x + 2 and y = –6x + 4, we need to find the values of x and y that satisfy both equations simultaneously. We can do this by setting the expressions for y in the two equations equal to each other, resulting in x² – 5x + 2 = –6x + 4. Solving this equation will give us the x-values, and substituting these x-values back into either of the original equations will yield the corresponding y-values.
10. Similar to question 9, to solve the system of equations y = x² + 8x + 2 and y = 7x + 4, we need to find the values of x and y that satisfy both equations.
11. To determine when the object will hit the ground in the equation h(t) = –16t² + 200, we need to find the value of t when h(t) = 0. This is because when the object hits the ground, its height will be 0. By setting h(t) = 0 and solving the resulting equation, we can determine the time it takes for the object to hit the ground.
12. To find how long it takes for the ball to reach its maximum height in the equation h(t) = –16t² + 32t + 6, we need to determine the t-value at which the function reaches its vertex. The vertex represents the maximum or minimum point of a quadratic function. Using the vertex formula t = -b / (2a), where a, b, and c are the coefficients in the equation, we can find the time it takes for the ball to reach its maximum height. Additionally, the maximum height can be found by substituting this t-value back into the equation to obtain the corresponding height.
13. Similar to question 12, to find how long it takes for the boulder to reach its maximum height in the equation h = –16t² + 148t + 30, we use the vertex formula. The maximum height is also found by substituting the t-value into the equation.
2. Shifted 5units down.
5. -7x^2 + 6x + 3 = 0
Use Quadratic formula and get 2 solutions : -0.354, and 1.211.
6. Y = 3x^2 + 18x + 27 = 0
x^2 * 6x + 9 = 0
(x+3)(x+3) = 0
x+3 = 0, x = -3.
x+3 = 0, X = -3.
2 solutions : -3, and -3.
11. Use Quad. Formula and get 3.54 s.
12. V = Vo + g*t = 0 @ max. ht.
32 - 32*t = 0
32t = 32
t = 1 s. = Time to reach max ht.
hmax = -16t^2 + 32t + 6
hmax = -16*1^2 + 32*1 + 6=22 Ft. after
1 second.
13. Same procedure as #12.