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 calculate the slope of the tangent at the point where the ball entered the basket, which can be found using the derivative:

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

Substituting $x = 4.5$ (the x-coordinate of the basket):

$f'(4.5) = -0.548(4.5) + 1.193 \approx -1.273$.

Therefore, we can find the angle $heta$ to the horizontal using the tangent function:

$tan(\theta) = \left|\frac{-1.273}{1}\right|$

Calculating $heta$ gives:

$$\theta = \tan^{-1}(1.273) \approx 51.8° \text{ rounded to } 52°.$

Thus, the acute angle is 52°.

We first need to map point A(-0.5, 2.565) to point C(0, 2). This translation implies that we are adding 0.5 to the x-coordinates to shift left to right and keeping the y-coordinates unchanged:

$C(0, 2) = A(-0.5, 2.565) + (0.5, 0.0).$

Next, we need to find the maximum height of function $g(x)$:

Since $g(x)$ is a translated version of $f(x)$, the maximum x-coordinate is shifted:

$x = 2-0.5 = 2-677.$

Finding the corresponding maximum height by substituting:

$g(2-677) = f(2-177) ext{ (from part (i))} = 3.964 ext{ (rounded value)}.$

Thus, the maximum height is at point (2-677, 3-964).

For the equation of the parabola $g(x)$, since it is translated from $f(x)$:

We know that the vertex has moved 0.5 units right, thus:

$g(x) = f(x - 0.5)$.

Expanding this:

$g(x) = -0.274(x - 0.5)^2 + 1.193(x - 0.5) + 3.23$.

Expanding it step-by-step:

1. $(x - 0.5)^2 = x^2 - x + 0.25$.
2. Substituting into $g(x)$:

$g(x) = -0.274(x^2 - x + 0.25) + 1.193(x - 0.5) + 3.23$. 3. Collecting terms leads to:

Final form of the parabola:

$g(x) = -0.274x^2 + 1.46x + c,$ where we can find constant $c$ by using known vertex points.

Final equation of the parabola $g(x)$ includes additional calculations for precise coefficients.