Circle the choices below that are quadratic
functions.
y=x² – 3
y=x² + 2x² – 4
y=x² + √x+7
y=10– x²

a) $169.2

b) $202.8

c) the church can bring either 31 or 34 members

Step-by-step explanation:

For each person the church brings that exceeds 15, the ticket price goes down by $0.20

18 members exceed 15 by 3.

This means that for 3 persons the church brings that exceeds 15, we have 3*0.20 = $0.60

The price of the ticket will now be

10 - 0.6 = $9.40

For the 18 members, we now have

18*9.4 = $169.2

b) The model for bringing a number of people at a particular cost = C(x)

x is the number of people that exceeds 15

C(x) = (15+x)(10-0.2x)

C(x) = 150 - 3x + 10x - 0.2x^2

= 150 + 7x - 0.2x^2

In bringing 26 people to the show, it is 11 more than 15

C(11) = 150 + 7(11) - 0.2(11)^2

C(11) = 150 +77 - 24.2

= $202.8

c) if the church can only afford $210.8, to solve x, we have

150 + 7x - 0.2x^2 = 210.8

-0.2x^2 + 7x + 150 = 210.8

-0.2x^2 + 7x + 150 - 210.8 = 0

-0.2x^2 + 7x -60.8 = 0

(-b +/-√b^2 - 4ac) / 2a

a = -0.2 , b = 7 c = -60.8

(-7 +/-√7^2 - 4(-0.2)(-60.8)) / 2(-0.2) = 0

(-7 +/-√49 - 48.64) / -0.4 = 0

(-7 +/-√0.36) / -0.4 = 0

(-7 +/- 0.6) / -0.4 = 0

(-7+0.6)/-0.4 or (-7 - 0.6)/-0.4

(-6.4)/-0.4 or (-7.6)/- 0.4

16 or 19

So we have (15+16) = 31 members

(15+19) = 34 members

The church can bring either 31 members or 34 members