1. Graph the quadratic equation y = x2 + x – 2. Identify the axis of symmetry and the vertex.
2. A ball is thrown from the top of a 50-ft building with an upward velocity of 24 ft/s. When will it reach its maximum height? How far above the ground will it be?
y = x^2 + x – 2
is y big if x big or big negative-->opens up, holds water
arrange to complete square
x^2 + x = y+2
x^2 + 1 x + (1/2)^2 = y + 2 + 1/4
(x + 1/2)^2 = y + 9/4
vertex at x = -1/2 and y = -9/4
axis of symmetry is x = -1/2
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most of us do physics in meters and stuff but anyway
in your units g = 32 ft/s^2
Hi = 50 ft
Vi = 24 ft/s
v = Vi - 32 t
v = 0 at top
32 t = 24
t = 3/4 = .75 seconds to top
h = Hi + Vi t - 16 t^2
h = 50 + 24 (3/4) -16 (9/16)
= 50 + 18 - 9
= 59 ft
1. http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+%2B+x+%E2%80%93+2
2. Use your favourite method to find the vertex of the parabola,
the x-coordinate tells you the "when", the y-coordinate tells you "how high"
1. axis of symmetry ... x = -b / 2a = -1 / 2
vertex is on the axis of symmetry ... y = (-1/2)^2 - 1/2 - 2 = -5/4
vertex is ... (-1/2 , -5/4)
2. h = -16 t^2 + 24 t + 50 ... max height is at the vertex (see #1)
Y=x2+4
Use the value of x=-3,-2,-1,1,2,3
Solve and graph in quadratic Equation
find the x intercepts of the graph y=x2-x-2
1. To graph the quadratic equation y = x^2 + x - 2, we can follow these steps:
Step 1: Set up a coordinate system by drawing x and y axes on a piece of graph paper.
Step 2: Plot some points to sketch the graph.
We can choose different values of x, calculate the corresponding y values, and plot these points on the graph. For example, let's choose x = -2, -1, 0, 1, and 2.
When x = -2, y = (-2)^2 + (-2) - 2 = 2. So we have the point (-2, 2).
When x = -1, y = (-1)^2 + (-1) - 2 = -2. So we have the point (-1, -2).
When x = 0, y = (0)^2 + (0) - 2 = -2. So we have the point (0, -2).
When x = 1, y = (1)^2 + (1) - 2 = 0. So we have the point (1, 0).
When x = 2, y = (2)^2 + (2) - 2 = 6. So we have the point (2, 6).
Step 3: Connect the plotted points to draw the graph.
In this case, the graph will be a parabola that opens upwards. Pass a smooth curve through the points to sketch the graph.
Step 4: Identify the axis of symmetry and the vertex.
The axis of symmetry is a vertical line that divides the graph into two symmetrical halves. It is given by the formula x = -b/2a, where a, b, and c are the coefficients of the quadratic equation. In our case, a = 1, b = 1, and c = -2. So, the axis of symmetry is x = -1/2.
The vertex of the parabola is the point on the graph where it reaches the highest or lowest point. The x-coordinate of the vertex is the same as the value of x on the axis of symmetry. In our case, x = -1/2. Plug this value into the equation y = x^2 + x - 2 to find the corresponding y-coordinate. So, y = (-1/2)^2 + (-1/2) - 2 = -2.75. Therefore, the vertex is (-1/2, -2.75).
2. To find when the ball will reach its maximum height and how far above the ground it will be, we need to use the principles of projectile motion.
Step 1: Find the time it takes for the ball to reach its maximum height.
The upward velocity of the ball is 24 ft/s, and the acceleration due to gravity is approximately -32 ft/s^2 (assuming upward as the positive direction). The ball will reach its maximum height when its vertical velocity becomes zero.
Using the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time, we can plug in the values:
0 = 24 - 32t
Simplifying the equation:
32t = 24
t = 24/32
t = 0.75 seconds
Therefore, the ball will reach its maximum height after 0.75 seconds.
Step 2: Calculate the maximum height above the ground.
To find the maximum height, we need to find the vertical displacement of the ball during this time.
Using the equation d = vit + 1/2at^2, where d is the vertical displacement, vi is the initial velocity, a is the acceleration, and t is the time, we can plug in the values:
d = (24)(0.75) + 1/2(-32)(0.75)^2
Simplifying the equation:
d = 18 - 9
d = 9 ft
Therefore, the ball will be 9 feet above the ground at its maximum height.
Damon is correct about the vertex
sorry about that...