Question 1 Part B (3 points): Using the GCF that you chose in part A, factor that GCF out of the polynomial:
1 answer:
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Answer:
- 168.25 ft
- 3.125 s
Step-by-step explanation:
It is almost too easy to let a graphing calculator show you the solution. (See attached)
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You may be expected to use some actual algebra (or even calculus) to find the solution.
If you've spent any time with quadratics, you know the vertex of ax²+bx+c is located at x=-b/(2a). Here, that means the time to get to the highest point is ...
... t = -100/(2·(-16)) = 100/32 = 3.125 . . . . seconds
Then the highest point is ...
... h(3.125) = -16·3.125² +100·3.125 +12 = 168.25 . . . . feet
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Or, you can put the equation into vertex form.
... h(t) = -16t² +100t +12
... = -16(t² +6.25t) +12
... = -16(t² +6.25t +3.125²) +12 +16·3.125² . . . . . add the square of half the t coefficient inside parentheses; add the opposite of the same amount outside parentheses
... = -16(t +3.125)² +168.25 . . . . . simplify
This tells us the peak of the travel of the bullet is at 168.25 ft after 3.125 seconds.
