Asked by Aria
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?
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?
Answers
Answered by
MathMate
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.
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.
Answered by
Anonymous
2.25
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