Answer:
1) Please find the attached drawing of the archway created with MS Excel
2) The y-intercept is (0, 24)
The x-intercepts are (-3, 0), and (8, 0)
3) The width of the archway at its base is 11
The height of the archway at its highest point = 30.25
Step-by-step explanation:
The given function representing the archway is y = -x² + 5·x + 24
1) Please find attached the required drawing of the archway created with MS Excel
2) The y-intercept is given by the point where x = 0
Therefore, we have, the y-value at the y-intercept = -0² + 5×0 + 24 = 24
The y-intercept = (0, 24)
The x-intercept is given by the point where y = 0
Therefore, the x-values at the x-intercept are found using the following equation;
0 = -x² + 5·x + 24
x² - 5·x - 24 = 0
By inspection, we have;
x² - 8·x + 3·x - 24 = 0
x·(x - 8) + 3·(x - 8) = 0
∴ (x + 3) × (x - 8) = 0
Either (x + 3) = 0, and x = -3, or (x - 8) = 0, and x = 8
Therefore, the x-intercepts are (-3, 0), and (8, 0)
3) The width of the archway at its base = The distance between the x-values at the two x-intercepts
∴ The width of the archway at its base = 8 - (-3) = 11
The highest point of the arch is given by the vertex of the parabola, y = a·x² + b·x + c, which has the x-value of the vertex = -b/(2·a)
∴ The x-value of the vertex of the given parabola, y = -x² + 5·x + 24, is x = -5/(2×(-1)) = 2.5
Therefore;
The y-value of the vertex, is y = -(2.5)² + 5×2.5 + 24 = 30.25 = The height of the archway at its highest point
∴ The height of the archway at its highest point = 30.25.
