Answer:
Answer is B.
Step-by-step explanation:
You can solve this by recognizing the vertex and the y-intercept of the quadratic equation based on the information provided.
The vertex occurs 6 seconds into the jump, when the cord is fully stretched to 72m, before it starts to contract again. This is when the jumper is closest to the ground, 122-72 = 50m above the ground. Notice the jumper never reaches the ground because the cord is only 72m long - it can't make it the full 122m - so there will not be an x-intercept (or t-intercept in this case). Anyway - the vertex is (6,50) using this information.
The y-intercept is the point at which t=0, the beginning of the jump. At the beginning of the jump, the jump is still on top of the cliff, 122m up. So the coordinates of the y-intercept are (0,122).
Now plug these values into the vertex form of the quadratic formula and solve for a:
[tex]f(t) = a(t-h)^2+k\\122=a(0-6)^2+50\\72 = 36a\\a=2[/tex]
This means our general equation (in vertex form) for this scenario is:
[tex]f(t)=2(t-6)^2+50[/tex]
Our answer options are in standard form, so just expand the vertex form of the equation to finish:
[tex]f(t)=2(t-6)^2+50\\f(t)=2(t^2-12t+36)+50\\f(t)=2t^2-24t+72+50\\f(t)=2t^2-24t+122[/tex]