How is the graph of y = 2x^2 + 4 different from the graph of y = 2x^2?

Question

How is the graph of y = 2x^2 + 4 different from the graph of y = 2x^2?

a. It is shifted 4 units up
b. It is shifted 4 units down
c. It is shifted 4 units right
d. It is shifted 4 units left

A ball is thrown into the air with an upward velocity of 48 ft/2. Its height h in feet after t seconds is given by the function h = −16t^2 + 48t + 5. How long does it take for the ball to reach its maximum height? What is the ball’s maximum height?
a. 1.5 seconds, 41 feet
b. 3 seconds, 5 feet
c. 1.5 seconds, 53 feet
d. 1.5 seconds, 113 feet

please help

Answer 1

For #1, picking a few ordered pairs, then plotting them will show the result.

#2, what method have you learned to find the vertex of this parabola?
Once you find the vertex, you have the time when the max is reached and its max height

Answer 2

For the first question:

The graph of y = 2x^2 + 4 is different from the graph of y = 2x^2 because it is shifted 4 units up. Therefore, the correct answer is option a. It is shifted 4 units up.

For the second question:
The equation for the height of the ball is h = −16t^2 + 48t + 5. To find the time it takes for the ball to reach its maximum height, we can use the formula t = -b/2a, where a, b, and c are coefficients of the quadratic equation in the form ax^2 + bx + c = 0.

In this case, the equation is h = −16t^2 + 48t + 5, so the coefficients are a = -16 and b = 48. Plugging these values into the formula, we get t = -48/(2*(-16)) = 48/32 = 1.5 seconds.

To find the maximum height, we need to substitute this value of t back into the equation for height. So, h = −16(1.5)^2 + 48(1.5) + 5 = -16(2.25) + 72 + 5 = -36 + 72 + 5 = 41 feet.

Therefore, the correct answer is option a. 1.5 seconds, 41 feet.

Answer 3

To determine the difference between the graph of y = 2x^2 + 4 and y = 2x^2, you can compare the two equations and observe the changes in the constants.

For y = 2x^2 + 4, the constant term is +4. This means the graph is shifted upward by 4 units compared to y = 2x^2. Therefore, the correct answer is a) It is shifted 4 units up.

Now let's move on to the next question about the ball's motion.

The equation h = -16t^2 + 48t + 5 describes the height of the ball after t seconds. To find when the ball reaches its maximum height, we need to determine the vertex of the parabolic function.

The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula:
x_vertex = -b / (2a)
y_vertex = f(x_vertex)

In this case, a = -16, b = 48, and c = 5. Plugging the values into the formula, we can find the x-coordinate of the vertex:

x_vertex = -48 / (2 * -16) = 48 / 32 = 3/2 or 1.5 seconds

To find the y-coordinate (the maximum height), we substitute x_vertex into the equation:

h = -(16 * (1.5)^2) + 48 * 1.5 + 5
h = -36 + 72 + 5
h = 41 feet

Therefore, the correct answer is a) 1.5 seconds, 41 feet.

How is the graph of y = 2x^2 + 4 different from the graph of y = 2x^2?

For #1, picking a few ordered pairs, then plotting them will show the result.

bruh.

For the first question:

To determine the difference between the graph of y = 2x^2 + 4 and y = 2x^2, you can compare the two equations and observe the changes in the constants.