Question

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?

Answer 1

To determine the attachment points of the banner to the archway, solve the system of equations formed by y = -x² + 6x and 4y = 21 − x. After arranging and equating the equations, apply the quadratic formula to find the x-values and then determine the corresponding y-values.

Step-by-step explanation:

To find the points where the banner, defined by the linear equation 4y = 21 − x, should be attached to the archway, modeled by the quadratic equation y = -x² + 6x, you need to solve a system of equations. The solutions will provide the x and y coordinates of the points where the banner crosses the archway.

First, rearrange the linear equation to make y the subject:

4y = 21 − x
y = (21 − x) / 4

Next, equate the two equations and solve for x:

-x² + 6x = (21 − x) / 4

Multiplying both sides by 4 to eliminate the fraction, we get:

-4x² + 24x = 21 − x
-4x² + 25x − 21 = 0

Now, solve the quadratic equation for x. You can do this either by factoring, completing the square, or using the quadratic formula. For simplicity, let's use the quadratic formula.

x = ∛(b² - 4ac) / 2a
where a = -4, b = 25, and c = -21

Substitute a, b, and c into the quadratic formula and calculate the values of x. These x-values will then be used to find the corresponding y-values by substituting them back into the equation y = -x² + 6x to find the two points where the banner should be attached.

Answer 2

Answer: (21/4, 63/16) and (1, 5).

Step-by-step explanation:

One possible method is to use substitution, as follows:

First, we isolate y in the second equation: 4y = 21 - x -> y = (21 - x) / 4

Then, we substitute y in the first equation with the expression we found: y = -x^2 + 6x -> (21 - x) / 4 = -x^2 + 6x

Next, we multiply both sides by 4 to eliminate the fraction: 21 - x = -4x^2 + 24x

Then, we rearrange the terms to get a quadratic equation in standard form: 4x^2 - 25x + 21 = 0

Finally, we solve the quadratic equation by factoring, using the quadratic formula, or completing the square2. One possible way is to factor, as follows:

We look for two numbers that multiply to get 4 * 21 = 84 and add to get -25. These numbers are -4 and -21.

We rewrite the equation as 4x^2 - 4x - 21x + 21 = 0

We group the terms and factor out the common factors: 4x(x - 1) - 21(x - 1) = 0

We factor out the common binomial factor: (4x - 21)(x - 1) = 0

We set each factor equal to zero and solve for x: 4x - 21 = 0 -> x = 21/4 and x - 1 = 0 -> x = 1

Now we have two values for x that satisfy the system of equations. To find the corresponding values for y, we plug them into either equation. For example, using the first equation, we get:

y = -x^2 + 6x -> y = -(21/4)^2 + 6(21/4) = 63/16

y = -x^2 + 6x -> y = -(1)^2 + 6(1) = 5

Therefore, the points where the banner should be attached to the archway are (21/4, 63/16) and (1, 5).