11. a) Sketch the following modulus functions and determine their domain and range i) $$ f(x)=|-3 x+1|+1 $$ ii) $$ f(x)=\left|x^{2}+5 x+4\right|-2 \quad $$ iv) $$ f(x)=\left|x^{2}+x-6\right|-3 $$ b) Given the quadratic function $$ f(x)-6 x^{2}+x \quad 2 $$ (i) Find the $$ y $$, intercept and the $$ x $$ - intercept. (ii) Determine the maximum or the minimum point of the function. (iii) Sketch the graph of $$ f(x)-6 x^{2}+x-2 $$ and the graph of $$ g(x)-\left|6 x^{2}+x-2\right| $$.

Question

11. a) Sketch the following modulus functions and determine their domain and range 	i) $$ f(x)=|-3 x+1|+1 $$ ii) $$ f(x)=\left|x^{2}+5 x+4\right|-2 \quad $$ iv) $$ f(x)=\left|x^{2}+x-6\right|-3 $$ b) Given the quadratic function $$ f(x)-6 x^{2}+x \quad 2 $$ (i) Find the $$ y $$, intercept and the $$ x $$ - intercept. (ii) Determine the maximum or the minimum point of the function. (iii) Sketch the graph of $$ f(x)-6 x^{2}+x-2 $$ and the graph of $$ g(x)-\left|6 x^{2}+x-2\right| $$.

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Let's break down the problem step by step.

### Part 11a: Sketching Modulus Functions

#### i) $$ f(x) = |-3x + 1| + 1 $$

1. **Domain**: The domain of $$ f(x) $$ is all real numbers, $$ (-\infty, \infty) $$.
2. **Range**: Since the absolute value function is always non-negative, the minimum value of $$ f(x) $$ occurs when $$ |-3x + 1| = 0 $$. This happens when $$ -3x + 1 = 0 $$ or $$ x = \frac{1}{3} $$. Thus, the minimum value of $$ f(x) $$ is $$ 1 $$. Therefore, the range is $$ [1, \infty) $$.

#### ii) $$ f(x) = |x^2 + 5x + 4| - 2 $$

1. **Domain**: The domain is all real numbers, $$ (-\infty, \infty) $$.
2. **Finding the roots of the quadratic**: 
   $$
   x^2 + 5x + 4 = 0
   $$
   Using the quadratic formula:
   $$
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - 16}}{2} = \frac{-5 \pm 3}{2}
   $$
   This gives us roots $$ x = -1 $$ and $$ x = -4 $$.
3. **Range**: The minimum value of $$ |x^2 + 5x + 4| $$ occurs at the vertex of the parabola, which is at $$ x = -\frac{b}{2a} = -\frac{5}{2} = -2.5 $$. Evaluating $$ f(-2.5) $$:
   $$
   f(-2.5) = |(-2.5)^2 + 5(-2.5) + 4| - 2 = |6.25 - 12.5 + 4| - 2 = |-2.25| - 2 = 2.25 - 2 = 0.25
   $$
   Therefore, the range is $$ [-2, \infty) $$.

#### iii) $$ f(x) = |x^2 + x - 6| - 3 $$

1. **Domain**: The domain is all real numbers, $$ (-\infty, \infty) $$.
2. **Finding the roots of the quadratic**:
   $$
   x^2 + x - 6 = 0
   $$
   Using the quadratic formula:
   $$
   x = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}
   $$
   This gives us roots $$ x = 2 $$ and $$ x = -3 $$.
3. **Range**: The vertex occurs at $$ x = -\frac{b}{2a} = -\frac{1}{2} $$. Evaluating $$ f(-0.5) $$:
   $$
   f(-0.5) = |(-0.5)^2 + (-0.5) - 6| - 3 = |0.25 - 0.5 - 6| - 3 = |-6.25| - 3 = 6.25 - 3 = 3.25
   $$
   Therefore, the range is $$ [-3, \infty) $$.

### Part 11b: Quadratic Function Analysis

Given the function $$ f(x) = -6x^2 + x - 2 $$:

#### i) Finding the intercepts

1. **Y-intercept**: Set $$ x = 0 $$:
   $$
   f(0) = -6(0)^2 + (0) - 2 = -2
   $$
   So, the y-intercept is $$ (0, -2) $$.

2. **X-intercepts**: Set $$ f(x) = 0 $$:
   $$
   -6x^2 + x - 2 = 0
   $$
   Using the quadratic formula:
   $$
   x = \frac{-1 \pm \sqrt{1 + 48}}{-12} = \frac{-1 \pm 7}{-12}
   $$
   This gives us:
   $$
   x = \frac{-8}{-12} = \frac{2}{3}, \quad x = \frac{6}{-12} = -\frac{1}{2}
   $$
   So, the x-intercepts are $$ \left(\frac{2}{3}, 0\right) $$ and $$ \left(-\frac{1}{2}, 0\right) $$.

#### ii) Maximum or Minimum Point

The vertex occurs at $$ x = -\frac{b}{2a} = -\frac{1}{-12} = \frac{1}{12} $$. Evaluating $$ f\left(\frac{1}{12}\right) $$:
$$
f\left(\frac{1}{12}\right) = -6\left(\frac{1}{12}\right)^2 + \left(\frac{1}{12}\right) - 2 = -\frac{6}{144} + \frac{1}{12} - 2 = -\frac{1}{24} + \frac{2}{24} - \frac{48}{24} = -\frac{47}{24}
$$
Thus, the maximum point is $$ \left(\frac{1}{12}, -\frac{47}{24}\right) $$.

#### iii) Sketching the Graphs

To sketch the graphs of $$ f(x) = -6x^2 + x - 2 $$ and $$ g(x) = |6x^2 + x - 2| $$, we can plot the quadratic function and its absolute value.

Now, I will proceed to calculate the necessary values for the graphs of $$ f(x) $$ and $$ g(x) $$.

### Calculating the Graphs

Let's calculate the vertex and intercepts for $$ g(x) = |6x^2 + x - 2| $$ to understand its shape.

1. **Finding the roots of $$ 6x^2 + x - 2 = 0 $$**:
   $$
   x = \frac{-1 \pm \sqrt{1 + 48}}{12} = \frac{-1 \pm 7}{12}
   $$
   This gives us roots $$ x = \frac{1}{3} $$ and $$ x = -\frac{2}{3} $$.

2. **Finding the vertex of $$ 6x^2 + x - 2 $$**
**Part 11a: Modulus Functions** 1. **i) $$ f(x) = |-3x + 1| + 1** - **Domain:** All real numbers - **Range:** $$ [1, \infty) $$ 2. **ii) $$ f(x) = |x^2 + 5x + 4| - 2** - **Domain:** All real numbers - **Range:** $$ [-2, \infty) $$ 3. **iii) $$ f(x) = |x^2 + x - 6| - 3** - **Domain:** All real numbers - **Range:** $$ [-3, \infty) $$ **Part 11b: Quadratic Function Analysis** 1. **i) Intercepts** - **Y-intercept:** $$ (0, -2) $$ - **X-intercepts:** $$ \left(\frac{2}{3}, 0\right) $$ and $$ \left(-\frac{1}{2}, 0\right) $$ 2. **ii) Maximum Point:** - Vertex at $$ \left(\frac{1}{12}, -\frac{47}{24}\right) $$ 3. **iii) Graphs of $$ f(x) = -6x^2 + x - 2 $$ and $$ g(x) = |6x^2 + x - 2| $$** - Plot the quadratic function and its absolute value to visualize the behavior of both functions.

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