Cohen has consistently averaged 120 yd. of horizontal distance with his 7-iron over the last two years playing golf. His ball flight typically reaches a maximum height of 24 yd. (Hint: A rough sketch will help you)
a.) Create a function in vertex form that relates these distances. Cohen is at the origin of the graph. Assume that the ball stops when it hits the ground.
b.) Determine the restrictions on the domain and range of this function.
c.) How high is the ball when it is 45 yd. from Cohen?
2.25
a) To create a function in vertex form that relates the distances, let's assume that the ball flight is represented by a parabolic path. In vertex form, the general equation is given by:
y = a(x - h)^2 + k
where (h, k) represents the vertex of the parabola.
In this case, Cohen's ball flight typically reaches a maximum height of 24 yards. Therefore, the vertex of the parabola is at (0, 24).
Also, Cohen has consistently averaged 120 yards of horizontal distance with his 7-iron. Since the horizontal distance is measured along the x-axis, we can set the x-coordinate of the vertex as 120:
(h, k) = (120, 24)
Plugging in these values into the vertex form equation, we get:
y = a(x - 120)^2 + 24
b) Now, let's determine the restrictions on the domain and range of this function:
- Domain: The horizontal distance, x, cannot be negative since it represents the distance traveled. Therefore, the domain is x >= 0.
- Range: The height, y, cannot be negative since the ball cannot be below the ground. Therefore, the range is y >= 0.
c) To find how high the ball is when it is 45 yards from Cohen, we can substitute x = 45 into the equation and solve for y:
y = a(45 - 120)^2 + 24
This will give us the height y at that particular horizontal distance.
To create a function in vertex form, let's start by looking at the given information and the shape of the golf ball's flight. From the given information, we know that Cohen's 7-iron consistently averages a horizontal distance of 120 yards and reaches a maximum height of 24 yards.
Based on this information, we can sketch a graph with the y-axis representing the height and the x-axis representing the horizontal distance. The shape of the graph would resemble a parabola opening downwards. The vertex of this parabola (the highest point) would be the maximum height of the ball.
Now, let's define the vertex form of a parabola:
y = a(x - h)^2 + k
Where (h, k) represents the vertex coordinates.
a) To create a function in vertex form, we need to find the values of a, h, and k. In this case, since the ball reaches a maximum height of 24 yards, the vertex would be (0, 24). Therefore, h = 0 and k = 24.
Now, let's find the value of a. To do this, we need another point on the parabola. We know that the horizontal distance is 120 yards when the ball hits the ground. Note that when the ball hits the ground, its height would be 0. Therefore, we can use the point (120, 0) to find the value of a.
Using the vertex form, we have:
0 = a(120 - 0)^2 + 24
Simplifying this equation, we get:
0 = 14400a + 24
Rearranging the equation, we find:
-14400a = 24
Dividing both sides by -14400, we get:
a = -24/14400 = -1/600
Therefore, the function in vertex form that relates these distances is:
y = (-1/600)x^2 + 24
b) Now let's determine the restrictions on the domain and range of this function.
Since the ball stops when it hits the ground, the height (y) cannot be negative. Therefore, the range of the function is [0, ∞).
The domain, on the other hand, is the set of all possible horizontal distances (x) that the ball can cover. In this case, there are no given restrictions on the horizontal distance, so the domain is (-∞, ∞).
c) To find how high the ball is when it is 45 yards from Cohen, we can substitute x = 45 into the equation and solve for y:
y = (-1/600)(45)^2 + 24
Simplifying this equation, we get:
y = -45^2/600 + 24
y = -2025/600 + 24
y = -3.375 + 24
y = 20.625
Therefore, when the ball is 45 yards from Cohen, its height is approximately 20.625 yards.
The standard quadratic equation in canonic form is
f(x)=y=a(x-h)+k
where (h,k) is the vertex.
when a<0, parabola is concave down.
So
for Cohen, h=120, k=24, so
f(x)=a(x-60)^2+24
However, we also know that y=0 at x=0 and x=120, so
f(x)=x(x-120)
Equate the two forms of f(x):
ax(x-120)≡a(x-60)^2+24
ax²-120ax≡ax²-120ax+3600a+24
Since if the two sides are equivalent, the coefficients of all powers are equal. This means:
3600a+24=0, or
a=-24/3600=-1/150
so
f(x)=-(a-60)^2/150+24
dom f(x)= R
range f(x) = (-∞,24]
I'll leave part (C) to you as an exercise.