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Big Ideas Math Algebra 2, 2014

BI

Big Ideas Math Algebra 2, 2014 View details

4. Modeling with Quadratic Functions

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Exercise 3 Page 77

Practice makes perfect

a The x-intercepts are still 4 and 24, however the y-intercept has changed from 9.6 to 4.8. This information is suited to use the intercept form of the equation.

y=a(x-p)(x-q)

y=a(x-p)(x-q)

p= 4, q= 24

y=a(x- 4)(x- 24)

x= 0, y= 4.8

4.8=a( 0-4)( 0-24)

Subtract terms

4.8=a(- 4)(- 24)

MultNegNegTwoPar

(- a)(- b)=a* b

4.8=96a

.LHS /96.=.RHS /96.

a=0.05

If we substitute a=0.05 into the intercept form equation along with p and q, we will get the function modeling the temperature in this case. y=0.05(x-4)(x-24) To find the minimum value we have to know the vertex of the parabola. Its x- coordinate always lies exactly in the middle between the x-coordinates of the x-intercepts. x=4+24/2=14 Now let's calculate the corresponding y-coordinate of the vertex.

f(x)=0.05(x-4)(x-24)

x= 14

f( 14)=0.05( 14-4)( 14-24)

Subtract terms

f(14)=0.05(10)(-10)

Multiply

f(14)=- 5

The coldest temperature is - 5^(∘)C at the same point of time as before, so after 14 hours.

b The parabola opens up and the axis of symmetry is x=14. Therefore, the function is decreasing over 0

Let's calculate the average rate of change over the interval 0

f(14)-f(0)/14-0

SubstitutePoints

Substitute ( 14,- 5) & ( 0,4.8)

- 5- 4.8/14-0

Subtract terms

- 9.8/14

Use a calculator

- 0.7

Now we will do the same for the second interval.

f(24)-f(14)/24-14

SubstitutePoints

Substitute ( 14,- 5) & ( 24,0)

0-( - 5)/24-14

Subtract terms

5/10

Calculate quotient

0.5

Since |- 0.7|>|0.5|, the average rate at which the temperature decreases is greater than the average rate at which temperature increases.

Subchapter links

Communicate Your Answer p.75

Monitoring Progress p.76-79

Exercises p.80-82