POINTS!! Draw a diagram of the archway modeled by the equation y = -x2 + 5x + 24. Find and label the y-intercept and the x-intercepts on the sketch. Then find and label the width of the archway at its base and the height of the archway at its highest point, assuming the base of the archway is along the x-axis.

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.

35-60 students are taking math.