Respuesta :
The first thing we must do in this case is to see which equations meet the following condition:
"A volleyball reaches its maximum height of 13 feet, 3 seconds after its served"
We have then:
Equation 1:
f (x) = - 2 (x-3) ^ 2 + 13
f (3) = - 2 (3-3) ^ 2 + 13
f (3) = 13
Yes, meet the condition
Equation 2:
f (x) = - 2x ^ 2 + 12x-5
f (3) = - 2 (3) ^ 2 + 12 (3) -5
f (3) = - 18 + 36-5
f (3) = 13
Yes, meet the condition
Answer:
B: f (x) = - 2x ^ 2 + 12x-5
D: f (x) = - 2 (x-3) ^ 2 + 13
"A volleyball reaches its maximum height of 13 feet, 3 seconds after its served"
We have then:
Equation 1:
f (x) = - 2 (x-3) ^ 2 + 13
f (3) = - 2 (3-3) ^ 2 + 13
f (3) = 13
Yes, meet the condition
Equation 2:
f (x) = - 2x ^ 2 + 12x-5
f (3) = - 2 (3) ^ 2 + 12 (3) -5
f (3) = - 18 + 36-5
f (3) = 13
Yes, meet the condition
Answer:
B: f (x) = - 2x ^ 2 + 12x-5
D: f (x) = - 2 (x-3) ^ 2 + 13
f(x)=a(x-h)^2+k
Maximum point: Vertex: V=(h,k)=(3,13)→h=3, k=13
Opens downward, then a<0 (negative)
f(x)=a(x-3)^2+13, with "a" negative
Possible option:
D: f(x)=-2(x-3)^2+13
Developing this expression:
f(x)=-2[(x)^2-2(x)(3)+(3)^2]+13
f(x)=-2(x^2-6x+9)+13
f(x)=-2x^2+12x-18+13
f(x)=-2x^2+12x-5. This is Option B
Answer: 2 Options:
Option B: f(x)=-2x^2+12x-5 and
Option D: f(x)=-2(x-3)^2+13
Maximum point: Vertex: V=(h,k)=(3,13)→h=3, k=13
Opens downward, then a<0 (negative)
f(x)=a(x-3)^2+13, with "a" negative
Possible option:
D: f(x)=-2(x-3)^2+13
Developing this expression:
f(x)=-2[(x)^2-2(x)(3)+(3)^2]+13
f(x)=-2(x^2-6x+9)+13
f(x)=-2x^2+12x-18+13
f(x)=-2x^2+12x-5. This is Option B
Answer: 2 Options:
Option B: f(x)=-2x^2+12x-5 and
Option D: f(x)=-2(x-3)^2+13