Help with parabola word problem

Question

While training for an upcoming event , Mary Jo made a dive whose path is described by the function f(x) = (5x^2/2 ) -10x+6 

 

a) sketch a graph of the equation. ( I already did this) 
b)from what height did she dive
c)how long did it take Mary Jo to reach the lowest point in her dive?

d)how long did she remain underwater?
e)how long did her dive last?
f)  determine the domaine and range of this function

 

In the bracket it's a fraction. 

 

Thank you for your help. 

Answer 1

b) From what height did she dive? ( Find Y-intercept for time at 0 seconds.)

 

f(0= 5/2 (0)^2-10(0) +6= 6 Y-intercept coordinate (0, 6)

The answer is 6 feet up if we are dealing with feet. Answer

c) Let time=x values We need to find her deepest swimming point, which is the minimum point and the vertex. Vertex =V(h,K), where h=X and k=y
Remember that f(x)= ax^2 + bX +C h= -b/2a and k=f(h)

F(x)= (5/2) X^2 –10X +6                 a=5/2      b = -10              c=6

Then use substitution: h= -b/2a = -(-10)/2(5/2)= 10/5 =2 seconds

F(2)= 5/2 (2)^2 –10(2) +6
5/2 (4)-20 + 6 = -4     v(2,-4) Answer

d) How long did she remain under water? Find the x-intercepts. Then subtract them, which will take away the distance from 0 to the first intercept. The time interval of the intercepts will give you her time under water. Use the quadratic formula, rather than factor.

0= 5/2 (X^2) –10X +6 a= 5/x b= -10 c=6

x = (10+ √40)/5 and x= (10 – √40)/5
x ≅ 3.26 seconds and x ≅ .7356 seconds

3.26-.74=2.52 seconds

e) How long did her dive last? This is from 0 seconds until she exited the water at the last intercept. 3.26 Answer

f) Domain (X values):   0 ≤ X ≤3.26

   Range ( Y values):       -4 ≤ y ≤0      If you like my work, please give me a thumbs up.

Answer 2

Alex, If you have already sketched the graph, look at what it is telling you in relation to the problem.  Mary Jo dives in at 0 seconds. (What is the height (Y) at this point)?  The vertex of the parabola is the lowest point of her dive.  The dive lasts as long as it takes from jump off (at x=0) till she surfaces at Y=0 at the second zero of the parabola.

 

Another question not asked:  How long is she in the water?  Think of the x-axis as the water line.