We are given the following:
- parabola passes to both (1,0) and (0,1)
<span> - slope at x = 1 is 4 from the equation of the tangent line </span>
<span>First, we figure out the value of c or the y intercept, we use the second point (0, 1) and substitute to the equation of the parabola. W</span><span>hen x = 0, y = 1. So, c should be equal to 1. The</span><span> parabola is y = ax^2 + bx + 1 </span>
<span>Now, we can substitute the point (1,0) into the equation,
</span>0 = a(1)^2 + b(1) + 1
<span>0 = a + b + 1
a + b = -1 </span>
<span>The slope at x = 1 is equal to 4 which is equal to the first derivative of the equation.</span>
<span>We take the derivative of the equation ,
y = ax^2 + bx + 1</span>
<span>y' = 2ax + b
</span>
<span>x = 1, y' = 2
</span>4 = 2a(1) + b
<span>4 = 2a + b </span>
So, we have two equations and two unknowns,<span> </span>
<span>2a + b = 4 </span>
<span>a + b = -1
</span><span>
Solving simultaneously,
a = 5 </span>
<span>b = -6</span>
<span>Therefore, the eqution of the parabola is y = 5x^2 - 6x + 1 .</span>
Answer:
C
You have to check which of numbers is the lowest.
Cheers :D
Wow !
OK. The line-up on the bench has two "zones" ...
-- One zone, consisting of exactly two people, the teacher and the difficult student.
Their identities don't change, and their arrangement doesn't change.
-- The other zone, consisting of the other 9 students.
They can line up in any possible way.
How many ways can you line up 9 students ?
The first one can be any one of 9. For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.
The total number of possible line-ups is
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .
But wait ! We're not done yet !
For each possible line-up, the teacher and the difficult student can sit
-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.
That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .
So the total total number of ways to do this is
(362,880) x (10) = 3,628,800 ways.
If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !
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