Answer: The banner must be 4.24 units long
a. Using the graphing method, find the points at which the equation of the angle of the banner will intersect the equation of the archway.
Sample answer:
Please see attached screenshot
Explanation:
The two equations intersect at (1, 5) and (4, 8), so the banner should be hung at these two points.
b. Using the algebraic method, confirm whether the two points obtained by the graphing method are correct.
Sample answer:
Given these two equations:
y = -x2 + 6x (1)
y – x = 4 (2)
Modify equation (2) to get y = x + 4.
To find the points of intersection, equate the two equations:
-x2 + 6x = x + 4
x2 - 5x + 4 = 0
(x – 4)(x – 1) = 0
x = 4 or x = 1.
Substitute the values of x in equation (2) to get the values of y:
y = x + 4
= (4) + 4
= 8.
y = x + 4
= (1) + 4
= 5
The two points of intersection are (4, 8) and (1, 5). These are the same points obtained by graphing.
c. Using the distance formula find the distance between points A and B. How long a banner does Ashton need to order?
Sample answer:
The distance formula is
x1 = 4
y1 = 8
x2 = 1
y2 = 5
Substituting these values in the formula:
d = 4.24 units.
The distance between the two points of intersection of the archway and the banner is 4.24 units, so the banner must be 4.24 units long.