Question

the archway of the main entrance of a university is modeled by the quadratic equation $y = -x^2 + 6x$. 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. (1.5,5.62) and (3.5,6.12)
b. (1,5.5) and (5.25,6.56)
c. (1.5,4.87) and (3.5,4.37)
d. (1,5) and (5.25,3.94)

Explanation:

Step1: Solve the second equation for y

From \(4y = 21 - x\), we get \(y=\frac{21 - x}{4}\).

Step2: Set the two equations equal

Set \(-x^{2}+6x=\frac{21 - x}{4}\). Multiply both sides by 4 to eliminate the denominator: \(-4x^{2}+24x = 21 - x\).

Step3: Rearrange into standard quadratic form

Bring all terms to one side: \(-4x^{2}+25x - 21 = 0\). Multiply both sides by -1: \(4x^{2}-25x + 21 = 0\).

Step4: Solve the quadratic equation

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here, \(a = 4\), \(b=-25\), \(c = 21\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-25)^{2}-4\times4\times21=625 - 336 = 289\).
Then, \(x=\frac{25\pm\sqrt{289}}{8}=\frac{25\pm17}{8}\).
This gives two solutions: \(x_1=\frac{25 + 17}{8}=\frac{42}{8}=5.25\) and \(x_2=\frac{25 - 17}{8}=\frac{8}{8}=1\).

Step5: Find the corresponding y - values

For \(x = 1\), substitute into \(y=\frac{21 - x}{4}\): \(y=\frac{21 - 1}{4}=\frac{20}{4}=5\).
For \(x = 5.25\), substitute into \(y=\frac{21 - x}{4}\): \(y=\frac{21 - 5.25}{4}=\frac{15.75}{4}=3.9375\approx3.94\).

Answer:

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