To determine the points at which the banner should be attached to the archway, we need to find the intersection points of the two equations given:
1. The equation of the archway: [tex]\( y = -x^2 + 6x \)[/tex]
2. The equation of the banner: [tex]\( 4y = 21 - x \)[/tex]
First, let's express the banner's equation in terms of [tex]\( y \)[/tex]:
[tex]\[ y = \frac{21 - x}{4} \][/tex]
Now we have two equations:
[tex]\[ y = -x^2 + 6x \][/tex]
[tex]\[ y = \frac{21 - x}{4} \][/tex]
To find the points of intersection, we set the two expressions for [tex]\( y \)[/tex] equal to each other:
[tex]\[ -x^2 + 6x = \frac{21 - x}{4} \][/tex]
To clear the fraction, multiply every term by 4:
[tex]\[ 4(-x^2 + 6x) = 21 - x \][/tex]
[tex]\[ -4x^2 + 24x = 21 - x \][/tex]
Rearrange the equation to set it to zero:
[tex]\[ -4x^2 + 24x + x - 21 = 0 \][/tex]
[tex]\[ -4x^2 + 25x - 21 = 0 \][/tex]
This is a quadratic equation. We can solve it using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -4 \)[/tex], [tex]\( b = 25 \)[/tex], and [tex]\( c = -21 \)[/tex]:
[tex]\[ x = \frac{-25 \pm \sqrt{25^2 - 4(-4)(-21)}}{2(-4)} \][/tex]
[tex]\[ x = \frac{-25 \pm \sqrt{625 - 336}}{-8} \][/tex]
[tex]\[ x = \frac{-25 \pm \sqrt{289}}{-8} \][/tex]
[tex]\[ x = \frac{-25 \pm 17}{-8} \][/tex]
This gives us two solutions for [tex]\( x \)[/tex]:
For the plus sign:
[tex]\[ x = \frac{-25 + 17}{-8} = \frac{-8}{-8} = 1 \][/tex]
For the minus sign:
[tex]\[ x = \frac{-25 - 17}{-8} = \frac{-42}{-8} = 5.25 \][/tex]
Now substitute [tex]\( x = 1 \)[/tex] and [tex]\( x = 5.25 \)[/tex] back into either original equation to find the corresponding [tex]\( y \)[/tex] values. Let's use [tex]\( y = -x^2 + 6x \)[/tex]:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -(1)^2 + 6(1) = -1 + 6 = 5 \][/tex]
For [tex]\( x = 5.25 \)[/tex]:
[tex]\[ y = -(5.25)^2 + 6(5.25) = -27.5625 + 31.5 = 3.9375 \][/tex]
Therefore, the points of intersection are:
[tex]\[ (1, 5) \text{ and } (5.25, 3.94) \][/tex]
The correct answer is:
D. [tex]\((1, 5) \text{ and } (5.25, 3.94)\)[/tex]
