The diagram shows Sarah’s first throw at the basket in a basketball game - Leaving Cert Mathematics - Question 8 - 2016

To find the angle of entry, we first need to calculate the derivative of the function to get the slope at the point where the ball enters the basket (x = 4):

$f'(x) = -0.548x + 1.193$

Substituting x = 4:

$f'(4) = -0.548(4) + 1.193 = -2.173$

The angle of the trajectory with respect to the horizontal can be calculated using the tangent function:

$\tan(\theta) = f'(4)$

Thus,

$\theta = \tan^{-1}(-2.173)$

Calculating this, we find:

$\theta \approx 51.8°$

Therefore, the acute angle to the horizontal is 52 degrees, rounded to the nearest degree.

Given the transformation of the graph, the mappings can be determined by:

• A = (-0.5, 2.565)
• C = (0, 2)

The transformation can be expressed as:

$C(0, 2) \text{ maps A to } (0.5, 2)$

From part (i), the maximum height previously calculated is 4.529 m. Adjusting this for the mapping:

From A to C, the vertical shift:

$h_{new} = h + (2 - 2.565) = 4.529 - 0.565 = 3.964$

At the translation, the new coordinates are around x = 2.677. Therefore, the centre of the ball reached its maximum height at (2.677, 3.964), correct to three decimal places.

From the transformations:

1. The parabola g(x) is similar to f(x) and can be expressed in the form:

since the vertex has changed, we follow:

$g(x) = a(x - h)^2 + k$

Using the new vertex (2.677, 3.964), the expression becomes:

$g(x) = a(x - 2.677)^2 + 3.964$

For the points (4, 3.05) in the context of g(x):

Substituting in:

$3.05 = a(4 - 2.677)^2 + 3.964$

Solving for 'a':

Once evaluated, we will find: $g(x) = -0.274(x - 2.677)^2 + 3.964$

Therefore, the equation of the parabola g(x) is converted back based on the transformations established.