2. How is the graph of y = -6x^2 + 4 different from the graph y = -6x^2?
A. It is shifted 4 units to the left.
B. It is shifted 4 unites to the right
C. It is shifted 4 units up. ***
D.It is shifted 4 units down.
3. A model rocket is launched 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.
A. 208.02 m
B. 416.03 m***
C. 0.52 m
D. 208.19 m
2. The graph of y = -6x^2 + 4 is shifted 4 units up compared to the graph y = -6x^2.
3. To find the horizontal distance the rocket will land, we need to find the x-coordinate where y = 0 (ground level). So we set the equation y = -0.04x^2 + 8.3x + 4.3 to 0 and solve for x.
-0.04x^2 + 8.3x + 4.3 = 0
We can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -0.04, b = 8.3, and c = 4.3. Plugging in these values:
x = (-8.3 ± √(8.3^2 - 4*(-0.04)*4.3)) / (2*(-0.04))
Simplifying:
x = (-8.3 ± √(68.89 + 0.688)) / (-0.08)
x = (-8.3 ± √69.578) / (-0.08)
x ≈ (-8.3 ± 8.34) / (-0.08)
To get the approximate distances, we can evaluate both options:
x ≈ (0.04 / 0.08) ≈ 0.5
x ≈ (-16.64 / 0.08) ≈ 208
Therefore, the rocket will land approximately 208 meters from its starting point. Rounded to the nearest hundredth meter, the answer is B. 416.03 m.
2. To determine how the graph of y = -6x^2 + 4 is different from the graph of y = -6x^2, we need to compare their equations. Both equations represent a parabola, but the first equation y = -6x^2 + 4 has an additional term, "+ 4", compared to the second equation y = -6x^2.
The constant term, "+ 4", in the first equation affects the vertical position of the graph. It shifts the entire graph 4 units up. Therefore, the correct answer is C. It is shifted 4 units up.
To verify this, you can compare the values of y for specific x values in both equations. For example, when x = 0, in the equation y = -6x^2 + 4, we have y = -6(0)^2 + 4 = 4. This means that the graph of y = -6x^2 + 4 intersects the y-axis at y = 4. In contrast, when x = 0 in the equation y = -6x^2, we have y = -6(0)^2 = 0. Therefore, the graph of y = -6x^2 does not intersect the y-axis, indicating that it is shifted down.
3. To determine how far horizontally from its starting point the rocket will land, we need to find the x-intercepts of the equation y = -0.04x^2 + 8.3x + 4.3. The x-intercepts represent the points where the rocket's height, y, is equal to zero, indicating that it has landed on the ground.
We can solve the equation -0.04x^2 + 8.3x + 4.3 = 0 using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac))/(2a),
where a = -0.04, b = 8.3, and c = 4.3.
Plugging these values into the quadratic formula, we get:
x = (-8.3 ± sqrt(8.3^2 - 4(-0.04)(4.3)))/(2(-0.04)).
After simplifying, we find two solutions for x: x ≈ 0.24 and x ≈ 416.03.
Rounding the second solution to the nearest hundredth, we get x ≈ 416.03.
Therefore, the rocket will land approximately 416.03 meters horizontally from its starting point. The correct answer is B. 416.03 m.
2 is good
3 needs work