Clueless on How to Solve this Word Problem

[Guest]

The downloading speed compared to 100 Mbps, y, during the hour of the day, x, is represented by the
graph y = 5x^2 −50x+ 252. For example, a time of 7:00 AM would be x =7. A time of 2:00 PM is x = 14.

a. Find at which time (measured in hh:mm) your downloading speed is equal to 100 Mbps.
b. What is the slowest time of day for downloading?
c. What is the slowest speed for downloading?
d. What download speed will be available at 9:30 PM?
e. What is the fastest speed for downloading?

[Guest]

Did you make no attempt to solve these yourself? They are pretty basic algebra problems.

a) Solve the quadratic equation [tex]5x^2- 50x+ 252= 100[/tex]. Subtract 100 from both sides to get the more standard form [tex]5x^2- 50x+ 152= 0[/tex]. That does not appear to factor easily so consider "completing the square" or the "quadratic formula".

b and c) "Completing the square" in the original [tex]5x^2- 50x+ 252[/tex] will let you write it as [tex]5(x- a)^2+ b[/tex] for specific numbers, a and b. Since a square is never negative, that has a smallest value of b when x= a.

d) 9:30 PM is 9.5+ 12= 21.5 hours into the day. Set x= 21.5 in [tex]5x^2- 50x+ 252[/tex] and do the calculation.

e) Since this is a parabola opening upward, the maximum value will be at one end of the work day. Since this appears to run all day, calculate [tex]5x^2- 50x+ 252[/tex] for x= 0 and x= 24 and choose the larger. (It seems peculiar to me that they are different since the end of one day IS the beginning of the next!)

[Guest]

Did you make no attempt to solve these yourself? They are pretty basic algebra problems.

a) Solve the quadratic equation [tex]5x^2- 50x+ 252= 100[/tex]. Subtract 100 from both sides to get the more standard form [tex]5x^2- 50x+ 152= 0[/tex]. That does not appear to factor easily so consider "completing the square" or the "quadratic formula".

No! Since y is "compared to 100 Mbps" which I take to mean "in 100s of Mbps" the equation would be [tex]5x^2- 50x+ 252= 1[/tex] so [tex]5x^2- 50x+ 251= 0[/tex]

b and c) "Completing the square" in the original [tex]5x^2- 50x+ 252[/tex] will let you write it as [tex]5(x- a)^2+ b[/tex] for specific numbers, a and b. Since a square is never negative, that has a smallest value of b when x= a.

d) 9:30 PM is 9.5+ 12= 21.5 hours into the day. Set x= 21.5 in [tex]5x^2- 50x+ 252[/tex] and do the calculation.

e) Since this is a parabola opening upward, the maximum value will be at one end of the work day. Since this appears to run all day, calculate [tex]5x^2- 50x+ 252[/tex] for x= 0 and x= 24 and choose the larger. (It seems peculiar to me that they are different since the end of one day IS the beginning of the next!)