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?

The point at which the banner should be attached to the archway is going to be given by the point(s) at which the two equations intercept:

y = -x^2 + 6x and 4y = 21 − x

Solving the system of equations, we find that they intercept at: (1, 5) and (5.25, 3.938)

Therefore, the banner should be attached to the archway at the points (1, 5) and (5.25, 3.938)

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!