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Sagot :
Let's solve the given problem step-by-step:
The problem is to find the points at which the banner should be attached to the archway. The archway is modeled by the quadratic equation:
[tex]\[ y = -x^2 + 6x \][/tex]
The equation of the line representing the angle at which the banner is hung is:
[tex]\[ 4y = 21 - x \][/tex]
First, we need to express both equations in a form that can be easily solved simultaneously.
Starting with the second equation:
[tex]\[ 4y = 21 - x \][/tex]
Let's solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{21 - x}{4} \][/tex]
Now, we have two equations:
1. [tex]\( y = -x^2 + 6x \)[/tex]
2. [tex]\( y = \frac{21 - x}{4} \)[/tex]
We can set these expressions for [tex]\( y \)[/tex] equal to each other:
[tex]\[ -x^2 + 6x = \frac{21 - x}{4} \][/tex]
Next, let's eliminate the fraction by multiplying through by 4:
[tex]\[ 4(-x^2 + 6x) = 21 - x \][/tex]
[tex]\[ -4x^2 + 24x = 21 - x \][/tex]
Bring all terms to one side of the equation to form a standard quadratic equation:
[tex]\[ -4x^2 + 24x + x - 21 = 0 \][/tex]
[tex]\[ -4x^2 + 25x - 21 = 0 \][/tex]
To solve this quadratic equation, we use 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].
Let's find the solutions for [tex]\( x \)[/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]:
[tex]\[ x_1 = \frac{-25 + 17}{-8} = \frac{-8}{-8} = 1 \][/tex]
[tex]\[ x_2 = \frac{-25 - 17}{-8} = \frac{-42}{-8} = 5.25 \][/tex]
Now, substitute these values back into one of the original equations to find the corresponding [tex]\( y \)[/tex]-values:
Using [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]
Thus, the points where the banner should be attached to the archway are:
[tex]\[ (1, 5) \][/tex]
[tex]\[ (5.25, 3.94) \][/tex]
The correct answer is C:
C. [tex]\( (1, 5) \)[/tex] and [tex]\( (5.25, 3.94) \)[/tex]
The problem is to find the points at which the banner should be attached to the archway. The archway is modeled by the quadratic equation:
[tex]\[ y = -x^2 + 6x \][/tex]
The equation of the line representing the angle at which the banner is hung is:
[tex]\[ 4y = 21 - x \][/tex]
First, we need to express both equations in a form that can be easily solved simultaneously.
Starting with the second equation:
[tex]\[ 4y = 21 - x \][/tex]
Let's solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{21 - x}{4} \][/tex]
Now, we have two equations:
1. [tex]\( y = -x^2 + 6x \)[/tex]
2. [tex]\( y = \frac{21 - x}{4} \)[/tex]
We can set these expressions for [tex]\( y \)[/tex] equal to each other:
[tex]\[ -x^2 + 6x = \frac{21 - x}{4} \][/tex]
Next, let's eliminate the fraction by multiplying through by 4:
[tex]\[ 4(-x^2 + 6x) = 21 - x \][/tex]
[tex]\[ -4x^2 + 24x = 21 - x \][/tex]
Bring all terms to one side of the equation to form a standard quadratic equation:
[tex]\[ -4x^2 + 24x + x - 21 = 0 \][/tex]
[tex]\[ -4x^2 + 25x - 21 = 0 \][/tex]
To solve this quadratic equation, we use 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].
Let's find the solutions for [tex]\( x \)[/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]:
[tex]\[ x_1 = \frac{-25 + 17}{-8} = \frac{-8}{-8} = 1 \][/tex]
[tex]\[ x_2 = \frac{-25 - 17}{-8} = \frac{-42}{-8} = 5.25 \][/tex]
Now, substitute these values back into one of the original equations to find the corresponding [tex]\( y \)[/tex]-values:
Using [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]
Thus, the points where the banner should be attached to the archway are:
[tex]\[ (1, 5) \][/tex]
[tex]\[ (5.25, 3.94) \][/tex]
The correct answer is C:
C. [tex]\( (1, 5) \)[/tex] and [tex]\( (5.25, 3.94) \)[/tex]
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