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The archway of the main entrance of a university is modeled by the quadratic equation \( y=-x^{2}+6 x \). The university is hanging a banner at the main entrance at an angle defined by the equation \( 4 y=21-x \). At what points should the banner be attached to the archway? A. \( (1.5,4.87) \) and \( (3.5,4.37) \) B. \( (1,5,5.62) \) and \( (3.5,6.12) \) C. \( (1,5) \) and \( (5.25,3.94) \) D. \( (1,5.5) \) and \( (5.25,6.56) \)

Ask by Martin Barnett. in the United States
Feb 04,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The banner should be attached at the points \( (1, 5) \) and \( (5.25, 3.94) \).

Solution

To determine where the banner should be attached to the archway, we need to find the intersection points between the archway and the banner. 1. **Equations Given:** - **Archway:** \( y = -x^2 + 6x \) - **Banner:** \( 4y = 21 - x \) which simplifies to \( y = -\frac{1}{4}x + \frac{21}{4} \) 2. **Find the Intersection Points:** Set the two equations equal to each other to find the \( x \)-coordinates where they intersect. \[ -x^2 + 6x = -\frac{1}{4}x + \frac{21}{4} \] Multiply both sides by 4 to eliminate the fraction: \[ -4x^2 + 24x = -x + 21 \] Bring all terms to one side: \[ -4x^2 + 25x - 21 = 0 \] Multiply by -1 for simplicity: \[ 4x^2 - 25x + 21 = 0 \] Solve for \( x \) using the quadratic formula: \[ x = \frac{25 \pm \sqrt{625 - 336}}{8} = \frac{25 \pm \sqrt{289}}{8} = \frac{25 \pm 17}{8} \] This gives two solutions: \[ x = \frac{42}{8} = 5.25 \quad \text{and} \quad x = \frac{8}{8} = 1 \] 3. **Find the Corresponding \( y \)-Coordinates:** Use the banner's equation \( y = -\frac{1}{4}x + \frac{21}{4} \). - For \( x = 1 \): \[ y = -\frac{1}{4}(1) + \frac{21}{4} = \frac{20}{4} = 5 \] - For \( x = 5.25 \): \[ y = -\frac{1}{4}(5.25) + \frac{21}{4} = -\frac{21}{16} + \frac{84}{16} = \frac{63}{16} \approx 3.94 \] 4. **Intersection Points:** The banner should be attached at \( (1, 5) \) and \( (5.25, 3.94) \). **Answer:** C. \( (1,5) \) and \( (5.25,3.94) \)

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Bonus Knowledge

To find the points where the banner should be attached to the archway, we must solve the system of equations represented by the quadratic equation \( y = -x^2 + 6x \) and the linear equation \( y = \frac{21 - x}{4} \). First, let's substitute the expression for \( y \) from the linear equation into the quadratic equation: \[ \frac{21 - x}{4} = -x^2 + 6x \] Multiplying both sides by 4 to eliminate the fraction gives: \[ 21 - x = -4x^2 + 24x \] Rearranging this results in the quadratic form: \[ 4x^2 - 25x + 21 = 0 \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 4, b = -25, c = 21 \). Calculating the discriminant: \[ b^2 - 4ac = (-25)^2 - 4 * 4 * 21 = 625 - 336 = 289 \] Now substituting into the quadratic formula: \[ x = \frac{25 \pm \sqrt{289}}{8} = \frac{25 \pm 17}{8} \] Calculating the two possible \( x \) values: 1. \( x = \frac{42}{8} = 5.25 \) 2. \( x = \frac{8}{8} = 1 \) Next, we find the corresponding \( y \)-values. Substituting \( x = 1 \) into the quadratic equation: \[ y = -1^2 + 6 * 1 = 5 \] For \( x = 5.25 \): \[ y = -5.25^2 + 6 * 5.25 = -27.5625 + 31.5 = 3.9375 \approx 3.94 \] Thus, the points where the banner should be attached to the archway are approximately \( (1, 5) \) and \( (5.25, 3.94) \). And this leads us to option C: \( (1,5) \) and \( (5.25,3.94) \). In order to prevent common mistakes while solving such equations, always ensure you clearly define one variable in terms of the other and watch out for algebraic errors, especially in simplification. If you want to dive deeper into quadratic equations and their applications, think about exploring topics like projectile motion in physics since it often models such parabolas beautifully. Understanding how these concepts can manifest in real-life scenarios adds an enriching layer to learning mathematics!

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