Calculate the value of the machine at the end of the third year if it depreciates at the rate of [tex]$15\%$[/tex] in the first year and [tex]$20\%$[/tex] yearly thereafter.

Copy and complete the table of values for the relation [tex]y = 2x - x^2, -2 \leq x \leq 4[/tex].
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -2 & -1.5 & -1 & -0.5 & 0 & 0.5 & 1 & 1.5 & 2 & 2.5 & 3 & 3.5 & 4 \\
\hline
[tex]$y$[/tex] & -8 & & -3 & & 0 & & 1 & & 0 & -1.3 & & -5.3 & \\
\hline
\end{tabular}

Using a scale of 2 cm to 1 unit on both axes, draw the graph for [tex]y = 2x - x^2, -2 \leq x \leq 4[/tex].

Using the graph, find the:
(i) maximum point;
(ii) line of symmetry;
(iii) range of values of [tex]$x$[/tex] for which [tex]y \geq 0[/tex].



Answer :

To solve the question, we need to follow a methodical approach, step by step:

### Step 1: Depreciation Calculation
We'll first determine the value of the machine at the end of the third year.

#### Calculation of Machine Depreciation:
- Original value of the machine: [tex]\( P \)[/tex]
- Depreciation rate in the first year: [tex]\( 15\% \)[/tex]
- Depreciation rate in the second and subsequent years: [tex]\( 20\% \)[/tex]

After the first year:
[tex]\[ \text{Value after 1st year} = P \times (1 - 0.15) = P \times 0.85 \][/tex]

After the second year:
[tex]\[ \text{Value after 2nd year} = P \times 0.85 \times (1 - 0.20) = P \times 0.85 \times 0.80 \][/tex]

After the third year:
[tex]\[ \text{Value after 3rd year} = P \times 0.85 \times 0.80 \times (1 - 0.20) = P \times 0.85 \times 0.80 \times 0.80 \][/tex]

So, at the end of the third year, the value of the machine will be:
[tex]\[ \text{Value} = P \times 0.85 \times 0.80 \times 0.80 \][/tex]
[tex]\[ = P \times 0.85 \times 0.64 \][/tex]
[tex]\[ = P \times 0.544 \][/tex]

The machine will be worth [tex]\( 54.4\% \)[/tex] of its original value.

### Step 2: Complete the Table for [tex]\( y = 2x - x^2 \)[/tex]
Now, we will compute the [tex]\( y \)[/tex] values for given [tex]\( x \)[/tex] values:

[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline x & -2 & -1.5 & -1 & -0.5 & 0 & 0.5 & 1 & 1.5 & 2 & 2.5 & 3 & 3.5 & 4 \\ \hline y & -8 & -5.25 & -3 & -1.25 & 0 & 0.75 & 1 & 0.75 & 0 & -1.25 & -3 & -5.25 & -8 \\ \hline \end{array} \][/tex]

### Step 3: Draw the Graph
You need to plot the graph of [tex]\( y = 2x - x^2 \)[/tex] using the given table values. Use a scale of 2 cm for 1 unit on both axes from [tex]\( x = -2 \)[/tex] to [tex]\( x = 4 \)[/tex].

### Step 4: Analyzing the Graph
From the graph, we can answer the specific questions:

(i) Maximum point:
From the completed table of values, we can see the maximum [tex]\( y \)[/tex] value is [tex]\( 1 \)[/tex] when [tex]\( x = 1 \)[/tex]. So, the maximum point is [tex]\( (1, 1) \)[/tex].

(ii) Line of symmetry:
For a quadratic function like [tex]\( y = 2x - x^2 \)[/tex], the line of symmetry is given by the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
In our equation, [tex]\( y = -x^2 + 2x \)[/tex], [tex]\( a = -1 \)[/tex] and [tex]\( b = 2 \)[/tex].
So, the line of symmetry is:
[tex]\[ x = -\frac{2}{2 \cdot (-1)} = 1 \][/tex]

(iii) Range of values of [tex]\( x \)[/tex] for which [tex]\( y \geq 0 \)[/tex]:
From the table, the values of [tex]\( x \)[/tex] for which [tex]\( y \geq 0 \)[/tex] are [tex]\( 0, 0.5, 1, 1.5, 2 \)[/tex].

So, summarizing:
1. Maximum point: [tex]\( (1, 1) \)[/tex]
2. Line of symmetry: [tex]\( x = 1 \)[/tex]
3. Range of [tex]\( x \)[/tex] values for which [tex]\( y \geq 0 \)[/tex]: [tex]\( [0, 2] \)[/tex]

I hope this detailed solution helps you understand the process and check your work!