The archway of the main entrance of a university is modeled by the quadratic equation
y = -x2 + 6x. The university is hanging a banner at the main entrance at an angle defined by the equation 4y = 21 − x. At what points should the banner be attached to the archway?
A.
(1,5.5) and (5.25,6.56)
B.
(1,5) and (5.25,3.94)
C.
(1.5,4.87) and (3.5,4.37)
D.
(1.5,5.62) and (3.5,6.12)

B. (1,5) and (5.25, 3.94)

Step-by-step explanation:

The answer is where the 2 equations intersect.

We need to solve the following system of equations:

y = -x^2 + 6x

4y = 21 - x

From the second equation:

x = 21 - 4y

Plug this into the first equation:

y = -(21 - 4y)^2 + 6(21 - 4y)

y = -(441 - 168y + 16y^2)+  126 - 24y

y = -441 + 168y - 16y^2 +  126 - 24y

16y^2 + y - 168y + 24y + 441 - 126 = 0

16y^2 - 143y + 315 = 0

y = [-(-143) +/- sqrt ((-143)^2 - 4 * 16 * 315)]/ (2*16)

y = 5, 3.938

When y = 5:

x = 21 - 4(5) = 1

When y = 3.938

x = 21 - 4(3.938) = 5.25.

1. 22

2. 1

3. 7

4. 18

step-by-step explanation

e2020 got it right