The point $$ (4,-91) $$ is a stationary point on the graph of $$ y=x^{3}-2 x^{2}+a x+c $$, where $$ c $$ is a constant. (i) Show that $$ a=-32 $$. (ii) Determine the value of $$ c $$. (iii) Find the coordinates of the other stationary point.

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To solve the problem, we will follow the steps outlined in the prompt.

### Step 1: Extract Known Conditions
We have the function:
$$
y = x^3 - 2x^2 + ax + c
$$
and a stationary point at $$ (4, -91) $$. A stationary point occurs where the derivative $$ \frac{dy}{dx} = 0 $$.

### Step 2: Find the Derivative
First, we need to find the derivative of the function:
$$
\frac{dy}{dx} = 3x^2 - 4x + a
$$

### Step 3: Substitute the Stationary Point into the Derivative
At the stationary point $$ x = 4 $$:
$$
3(4)^2 - 4(4) + a = 0
$$
Calculating this gives:
$$
3(16) - 16 + a = 0 \
48 - 16 + a = 0 \
32 + a = 0
$$
Thus, we find:
$$
a = -32
$$

### Step 4: Show that $$ a = -32 $$
We have shown that $$ a = -32 $$.

### Step 5: Determine the Value of $$ c $$
Next, we substitute the coordinates of the stationary point $$ (4, -91) $$ into the original function to find $$ c $$:
$$
-91 = (4)^3 - 2(4)^2 + (-32)(4) + c
$$
Calculating this gives:
$$
-91 = 64 - 32 - 128 + c \
-91 = -96 + c \
c = -91 + 96 \
c = 5
$$

### Step 6: Find the Coordinates of the Other Stationary Point
Now we have $$ a = -32 $$ and $$ c = 5 $$. The function becomes:
$$
y = x^3 - 2x^2 - 32x + 5
$$
We need to find the stationary points by setting the derivative to zero:
$$
3x^2 - 4x - 32 = 0
$$
We will solve this quadratic equation using the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where $$ a = 3 $$, $$ b = -4 $$, and $$ c = -32 $$.

Calculating the discriminant:
$$
b^2 - 4ac = (-4)^2 - 4(3)(-32) = 16 + 384 = 400
$$
Now substituting into the quadratic formula:
$$
x = \frac{4 \pm \sqrt{400}}{2(3)} = \frac{4 \pm 20}{6}
$$
Calculating the two possible values for $$ x $$:
1. $$ x = \frac{24}{6} = 4 $$ (the known stationary point)
2. $$ x = \frac{-16}{6} = -\frac{8}{3} $$

### Step 7: Find the Corresponding $$ y $$ Value
Now we find the $$ y $$ value for $$ x = -\frac{8}{3} $$:
$$
y = \left(-\frac{8}{3}\right)^3 - 2\left(-\frac{8}{3}\right)^2 - 32\left(-\frac{8}{3}\right) + 5
$$
Calculating each term:
1. $$ \left(-\frac{8}{3}\right)^3 = -\frac{512}{27} $$
2. $$ -2\left(-\frac{8}{3}\right)^2 = -2 \cdot \frac{64}{9} = -\frac{128}{9} $$
3. $$ -32\left(-\frac{8}{3}\right) = \frac{256}{3} $$

Now, we need a common denominator to combine these:
- The common denominator is 27.

Converting each term:
1. $$ -\frac{512}{27} $$
2. $$ -\frac{128}{9} = -\frac{384}{27} $$
3. $$ \frac{256}{3} = \frac{2304}{27} $$
4. $$ 5 = \frac{135}{27} $$

Now summing these:
$$
y = -\frac{512}{27} - \frac{384}{27} + \frac{2304}{27} + \frac{135}{27} = \frac{2304 - 512 - 384 + 135}{27} = \frac{1543}{27}
$$

### Final Answer
The coordinates of the other stationary point are:
$$
\left(-\frac{8}{3}, \frac{1543}{27}\right)
$$

### Summary of Results
(i) $$ a = -32 $$  
(ii) $$ c = 5 $$  
(iii) The coordinates of the other stationary point are $$ \left(-\frac{8}{3}, \frac{1543}{27}\right) $$.
$$ a = -32 $$, $$ c = 5 $$, and the other stationary point is at $$ \left(-\frac{8}{3}, \frac{1543}{27}\right) $$.

The point \( (4,-91) \) is a stationary point on the graph of \( y=x^{3}-2 x^{2}+a x+c \), where \( c \) is a constant. (i) Show that \( a=-32 \). (ii) Determine the value of \( c \). (iii) Find the coordinates of the other stationary point.

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