in graphing quadratic functions what are the ordered pairs for y=x*x+2
describe how you can use the characteristics of quadratic functions to determine the ordered pairs of 5 distinct points on the quadratic function with the following characteristics:
- zero at x = -4
- axis of symmetry at x = 2
- max value of 9
- y-intercept at 8
To determine the ordered pairs of 5 distinct points on the quadratic function with the given characteristics, we can use the general form of a quadratic function: f(x) = a(x - h)^2 + k.
1. Zero at x = -4:
Since x = -4 is a zero of the quadratic function, we know that when x = -4, y = 0. Therefore, one ordered pair is (-4, 0).
2. Axis of symmetry at x = 2:
The axis of symmetry is the vertical line that passes through the vertex of the quadratic function. Since the axis of symmetry is at x = 2, we know that the vertex of the quadratic function is at (2, k), where k is the y-value of the vertex.
3. Max value of 9:
Since the max value is at the vertex, we know that the y-value of the vertex is 9. So the vertex is (2, 9).
4. Y-intercept at 8:
The y-intercept is the point where the graph crosses or intersects the y-axis. Since the y-intercept is at (0, 8), we know that when x = 0, y = 8.
5. Finding the additional ordered pairs:
To find three more distinct points, we can choose x-values that are equidistant from both sides of the vertex. For example, we can choose x = 1 and x = 3.
When x = 1:
f(1) = a(1 - 2)^2 + 9
f(1) = a(-1)^2 + 9
f(1) = a + 9
When x = 3:
f(3) = a(3 - 2)^2 + 9
f(3) = a(1)^2 + 9
f(3) = a + 9
Since both equations give us f(1) = a + 9 and f(3) = a + 9, we can say that f(1) = f(3).
Therefore, additional ordered pairs are (1, f(1)) and (3, f(3)).
To summarize, the 5 distinct ordered pairs on the quadratic function are:
(-4, 0)
(0, 8)
(1, f(1))
(2, 9)
(3, f(3))
how do I write on the graph the summarize of the 5 distinct ordered pairs on the quadratic function are:
(-4, 0)
(0, 8)
(1, f(1))
(2, 9)
(3, f(3))
To write the summary of the 5 distinct ordered pairs on the graph, you can label the points directly on the graph.
1. The point (-4, 0) represents the zero of the quadratic function and can be labeled as a point on the x-axis at x = -4 with a y-coordinate of 0.
2. The point (0, 8) represents the y-intercept of the quadratic function and can be labeled as a point on the y-axis at y = 8 with an x-coordinate of 0.
3. The point (1, f(1)) represents one of the additional distinct points. To find the value of f(1), substitute x = 1 into the quadratic function. For example, let's say f(x) = ax^2 + bx + c. When x = 1, the ordered pair becomes (1, a(1)^2 + b(1) + c) or (1, a + b + c). So, you'll need to know the values of a, b, and c in the quadratic function to determine the exact y-coordinate.
4. The point (2, 9) represents the vertex of the quadratic function and can be labeled as a point on the graph at x = 2 with a y-coordinate of 9.
5. The point (3, f(3)) represents another one of the additional distinct points. Similarly to point 3, you'll need to substitute x = 3 into the quadratic function to determine the y-coordinate.
After obtaining the values for f(1) and f(3), you can label the points (1, f(1)) and (3, f(3)) on the graph.
Make sure to label the points clearly and accurately on the graph to represent the 5 distinct ordered pairs.
an object on earth falls with an acceleration of a = 9.81 m/s^2. a function exists between the height form which the object falls, the initial velocity of the object, and the time the object spends in the air.
h(t) = Vt + at^2
V = initial velocity
t = time
a = acceleration due to gravity
lucy and Kim determine the height of a building by dropping a brick from its roof. the initial velocity will be 0 m/s because lucy will simply let go of the brick. Kim times the fall of the brick to be 2.75 seconds.
- determine the height of the building to the nearest hundredth?
To determine the height of the building using the given information, we can use the formula h(t) = Vt + at^2, where h(t) represents the height, V represents the initial velocity (which is 0 m/s in this case), t represents the time, and a represents the acceleration due to gravity (which is approximately 9.81 m/s^2).
Substituting the given values into the formula, we have:
h(t) = 0*t + 9.81*t^2
h(t) = 9.81*t^2
Since we are given that Kim times the fall of the brick to be 2.75 seconds (t = 2.75), we can substitute this value into the equation to find the height:
h(2.75) = 9.81*(2.75)^2
h(2.75) ≈ 9.81*(7.5625)
h(2.75) ≈ 73.97906
Therefore, the height of the building, to the nearest hundredth, is approximately 73.98 meters.
an object on earth falls with an acceleration of a = 9.81 m/s^2. a function exists between the height form which the object falls, the initial velocity of the object, and the time the object spends in the air.
h(t) = Vt + at^2
V = initial velocity
t = time
a = acceleration due to gravity
lucy and Kim determine the height of a building by dropping a brick from its roof. the initial velocity will be 0 m/s because lucy will simply let go of the brick. Kim times the fall of the brick to be 2.75 seconds.
- determine the amount of time the brick is in the air if lucy throws the brick downward with an initial velocity of 4.25 m/s. round to the nearest hundredth of a second.
To determine the time the brick is in the air when Lucy throws the brick downward with an initial velocity of 4.25 m/s, we can use the same formula h(t) = Vt + at^2.
Since Lucy throws the brick downward, the initial velocity (V) will be -4.25 m/s (negative because it is directed downward).
Let's set h(t) equal to zero, since we want to find the time when the height is zero (when the brick touches the ground):
0 = -4.25t + 9.81t^2
To solve for t, we need to rearrange the equation into quadratic form:
9.81t^2 - 4.25t = 0
Now, we can factor out t from both terms:
t(9.81t - 4.25) = 0
So, either t = 0 or (9.81t - 4.25) = 0
Solving the equation 9.81t - 4.25 = 0 for t, we have:
9.81t = 4.25
t ≈ 4.25 / 9.81
t ≈ 0.4341
Therefore, the brick is in the air for approximately 0.43 seconds when Lucy throws the brick downward with an initial velocity of 4.25 m/s.
To find the ordered pairs for the quadratic function y = x^2 + 2, we need to substitute different values of x into the equation and calculate the corresponding values of y.
Let's choose a few values for x and calculate the corresponding y-values:
1. When x = -2:
y = (-2)^2 + 2
= 4 + 2
= 6
So, one ordered pair is (-2, 6).
2. When x = -1:
y = (-1)^2 + 2
= 1 + 2
= 3
The second ordered pair is (-1, 3).
3. When x = 0:
y = (0)^2 + 2
= 0 + 2
= 2
The third ordered pair is (0, 2).
4. When x = 1:
y = (1)^2 + 2
= 1 + 2
= 3
The fourth ordered pair is (1, 3).
5. When x = 2:
y = (2)^2 + 2
= 4 + 2
= 6
The fifth ordered pair is (2, 6).
Therefore, the ordered pairs for the quadratic function y = x^2 + 2 are (-2, 6), (-1, 3), (0, 2), (1, 3), and (2, 6).