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

The angle at which the ball enters the basket can be found using the derivative of the function at point A.

First, we calculate the derivative: $f'(x) = -0.548x + 1.193$

Now we evaluate the derivative at $A(x = -0.5)$: $f'(-0.5) = -0.548(-0.5) + 1.193 \approx 1.467$

Using the tangent function, we find the angle of entry: $\tan(\theta) = 1.467$

Thus, $\theta = \tan^{-1}(1.467) \approx 56.1^{\circ}$

Rounding to the nearest degree, the acute angle is 56 degrees.

The translation from point A to point C involves moving A(-0.5, 2.565) to C(0, 2). This translates as:

• For $x$: The shift is by +0.5.
• For $y$: The shift is by +1.435 (since $3.23-2.565 = 0.665$, and $3.23+1.435$ gives the new height).

Thus, we find that the new maximum point after translation is: $g(2.677) = f(2.177) + 1.435 \approx 4.529 - 0.565 \approx 3.964$

Hence, the maximum height reached by the ball after translation corresponds to the point (2.677, 3.964).

The equation of the parabola $g(x)$ can be derived from $f(x)$ using the transformations to point C(0, 2).

Starting with $f(x)$: $f(x) = -0.274x^2 + 1.193x + 3.23$

Now, moving A(-0.5) to C(0), we compute:

• The changed vertex from (2.177, 4.529) using the vertex form: $g(x) = a(x - h)^2 + k$

Using:

• $h = 2.677$,
• $k = 3.964$.

Solving will yield: $g(x) = -0.274(x - 2.677)^2 + 3.964$

This represents the equation of the parabola after applying the transformation.