Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
4. Modeling with Quadratic Functions
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Exercise 33 Page 82

Practice makes perfect
a We are given a parabola modeling the number of students absent due to the flu each day.
The vertex lies at 6,19. It means that on the 6th day the number of absent students reached a maximum equal to 19.
b Since we are given the vertex (6,19), it is reasonable to use the vertex form of the equation of a parabola.
y=a(x-h)^2+k Here h is the x-coordinate and k is the y-coordinate of the vertex. y=a(x- 6)^2+ 19 To complete the equation we need to substitute the second point (0,1) into the equation and calculate a.
y=a(x-6)^2+19
1=a( 0-6)^2+19
1=a(- 6)^2+19
1=36a+19
- 18=36a
a=- 0.5
Let's complete the equation by substituting a=- 0.5 into it. y=- 0.5(x-6)^2+19 Now, we can estimate the number of absent students on the 10th day.
f(x)=- 0.5(x-6)^2+19
f( 10)=- 0.5( 10-6)^2+19
f(10)=- 0.5(4)^2+19
f(10)=- 0.5(16)+19
f(10)=- 8+19
f(10)=11
On the 10th day there should be about 11 students absent.
c The domain of the parabola is divided into two intervals. The function increases over 0Average Rate of Change over this interval.

f(6)-f(0)/6-0=19-1/6=3 The function decreases over 6|- 2.5|, the average rate at which the number of absent students increases is greater than the average rate at which it decreases.