Question

The archway of the main entrance of a university is modeled by the quadratic equation

Answer 1

Answer:

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.