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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 = βˆ’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

bruh.

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 = βˆ’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 = βˆ’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 = βˆ’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.

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.

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