Answer :
Certainly! Let's work through each part of the question step-by-step.
### Part (a): Evaluate [tex]\( f(1) \)[/tex] and hence state a linear factor of [tex]\( f(x) \)[/tex]
First, let's evaluate the function [tex]\( f(x) \)[/tex] at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^3 - 9 \cdot 1^2 + 24 \cdot 1 - 16 \][/tex]
Calculating each term:
[tex]\[ 1^3 = 1 \][/tex]
[tex]\[ 9 \cdot 1^2 = 9 \][/tex]
[tex]\[ 24 \cdot 1 = 24 \][/tex]
[tex]\[ f(1) = 1 - 9 + 24 - 16 = 0 \][/tex]
Since [tex]\( f(1) = 0 \)[/tex], [tex]\( x = 1 \)[/tex] is a root of [tex]\( f(x) \)[/tex]. Therefore, [tex]\( (x - 1) \)[/tex] is a linear factor of [tex]\( f(x) \)[/tex].
### Part (b): Show that [tex]\( f(x) \)[/tex] can be expressed in the form [tex]\( f(x) = (x + p)(x + q)^2 \)[/tex], where [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are integers to be found.
Given that [tex]\( (x - 1) \)[/tex] is a factor, we perform polynomial division of [tex]\( f(x) \)[/tex] by [tex]\( (x - 1) \)[/tex] to find the quotient.
The polynomial division results in a quotient and remainder. Denote the polynomial by:
[tex]\[ f(x) = x^3 - 9x^2 + 24x - 16 \][/tex]
and perform the division by [tex]\( x - 1 \)[/tex]:
The quotient is:
[tex]\[ x^2 - 8x + 16 \][/tex]
and the remainder is:
[tex]\[ 0 \][/tex]
Next, factorize the quotient [tex]\( x^2 - 8x + 16 \)[/tex]:
[tex]\[ x^2 - 8x + 16 = (x - 4)^2 \][/tex]
Thus, we can express [tex]\( f(x) \)[/tex] as:
[tex]\[ f(x) = (x - 1)(x - 4)^2 \][/tex]
To match it with the form [tex]\( (x + p)(x + q)^2 \)[/tex]:
[tex]\[ (x - 1) = (x + p) \Rightarrow p = -1 \][/tex]
[tex]\[ (x - 4) = (x + q) \Rightarrow q = -4 \][/tex]
Hence, [tex]\( f(x) = (x - 1)(x - 4)^2 = (x + (-1))(x + (-4))^2 \)[/tex].
### Part (c): Sketch the curve [tex]\( y = f(x) \)[/tex]
To sketch the curve, we need to identify key features:
- Roots: From the factorized form [tex]\( f(x) = (x - 1)(x - 4)^2 \)[/tex], the roots are [tex]\( x = 1 \)[/tex] and [tex]\( x = 4 \)[/tex].
- Behavior around roots:
- The root at [tex]\( x = 1 \)[/tex] is a simple root, so the curve passes through [tex]\( x = 1 \)[/tex].
- The root at [tex]\( x = 4 \)[/tex] is a double root, so the curve touches the x-axis at [tex]\( x = 4 \)[/tex] and turns around.
- End behavior: Since the highest degree term in [tex]\( f(x) \)[/tex] is [tex]\( x^3 \)[/tex], as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex], and as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
Here’s the envisaged sketch of the curve [tex]\( y = f(x) \)[/tex]:
1. Plot the roots at [tex]\( x = 1 \)[/tex] and [tex]\( x = 4 \)[/tex].
2. The curve crosses the x-axis at [tex]\( x = 1 \)[/tex].
3. The curve touches the x-axis and turns back at [tex]\( x = 4 \)[/tex].
Visual Representation:
```
^
|
5 | /\
| / \ (double root at x=4)
0 |--------(1)-----(4)----------
| | /
-5 | | /
| _/\ /
| / /
-|----/------/------>
```
The curve will pass through [tex]\( x = 1 \)[/tex] and will touch and turn around at [tex]\( x = 4 \)[/tex], forming a local minimum or maximum depending on the derivative at those points.
### In Summary
- Evaluated [tex]\( f(1) = 0 \)[/tex], thus [tex]\( x - 1 \)[/tex] is a linear factor.
- Expressed [tex]\( f(x) \)[/tex] as [tex]\( (x - 1)(x - 4)^2 \)[/tex], so [tex]\( p = -1 \)[/tex] and [tex]\( q = -4 \)[/tex].
- Sketched the curve based on roots and the behavior around them.
### Part (a): Evaluate [tex]\( f(1) \)[/tex] and hence state a linear factor of [tex]\( f(x) \)[/tex]
First, let's evaluate the function [tex]\( f(x) \)[/tex] at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^3 - 9 \cdot 1^2 + 24 \cdot 1 - 16 \][/tex]
Calculating each term:
[tex]\[ 1^3 = 1 \][/tex]
[tex]\[ 9 \cdot 1^2 = 9 \][/tex]
[tex]\[ 24 \cdot 1 = 24 \][/tex]
[tex]\[ f(1) = 1 - 9 + 24 - 16 = 0 \][/tex]
Since [tex]\( f(1) = 0 \)[/tex], [tex]\( x = 1 \)[/tex] is a root of [tex]\( f(x) \)[/tex]. Therefore, [tex]\( (x - 1) \)[/tex] is a linear factor of [tex]\( f(x) \)[/tex].
### Part (b): Show that [tex]\( f(x) \)[/tex] can be expressed in the form [tex]\( f(x) = (x + p)(x + q)^2 \)[/tex], where [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are integers to be found.
Given that [tex]\( (x - 1) \)[/tex] is a factor, we perform polynomial division of [tex]\( f(x) \)[/tex] by [tex]\( (x - 1) \)[/tex] to find the quotient.
The polynomial division results in a quotient and remainder. Denote the polynomial by:
[tex]\[ f(x) = x^3 - 9x^2 + 24x - 16 \][/tex]
and perform the division by [tex]\( x - 1 \)[/tex]:
The quotient is:
[tex]\[ x^2 - 8x + 16 \][/tex]
and the remainder is:
[tex]\[ 0 \][/tex]
Next, factorize the quotient [tex]\( x^2 - 8x + 16 \)[/tex]:
[tex]\[ x^2 - 8x + 16 = (x - 4)^2 \][/tex]
Thus, we can express [tex]\( f(x) \)[/tex] as:
[tex]\[ f(x) = (x - 1)(x - 4)^2 \][/tex]
To match it with the form [tex]\( (x + p)(x + q)^2 \)[/tex]:
[tex]\[ (x - 1) = (x + p) \Rightarrow p = -1 \][/tex]
[tex]\[ (x - 4) = (x + q) \Rightarrow q = -4 \][/tex]
Hence, [tex]\( f(x) = (x - 1)(x - 4)^2 = (x + (-1))(x + (-4))^2 \)[/tex].
### Part (c): Sketch the curve [tex]\( y = f(x) \)[/tex]
To sketch the curve, we need to identify key features:
- Roots: From the factorized form [tex]\( f(x) = (x - 1)(x - 4)^2 \)[/tex], the roots are [tex]\( x = 1 \)[/tex] and [tex]\( x = 4 \)[/tex].
- Behavior around roots:
- The root at [tex]\( x = 1 \)[/tex] is a simple root, so the curve passes through [tex]\( x = 1 \)[/tex].
- The root at [tex]\( x = 4 \)[/tex] is a double root, so the curve touches the x-axis at [tex]\( x = 4 \)[/tex] and turns around.
- End behavior: Since the highest degree term in [tex]\( f(x) \)[/tex] is [tex]\( x^3 \)[/tex], as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex], and as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
Here’s the envisaged sketch of the curve [tex]\( y = f(x) \)[/tex]:
1. Plot the roots at [tex]\( x = 1 \)[/tex] and [tex]\( x = 4 \)[/tex].
2. The curve crosses the x-axis at [tex]\( x = 1 \)[/tex].
3. The curve touches the x-axis and turns back at [tex]\( x = 4 \)[/tex].
Visual Representation:
```
^
|
5 | /\
| / \ (double root at x=4)
0 |--------(1)-----(4)----------
| | /
-5 | | /
| _/\ /
| / /
-|----/------/------>
```
The curve will pass through [tex]\( x = 1 \)[/tex] and will touch and turn around at [tex]\( x = 4 \)[/tex], forming a local minimum or maximum depending on the derivative at those points.
### In Summary
- Evaluated [tex]\( f(1) = 0 \)[/tex], thus [tex]\( x - 1 \)[/tex] is a linear factor.
- Expressed [tex]\( f(x) \)[/tex] as [tex]\( (x - 1)(x - 4)^2 \)[/tex], so [tex]\( p = -1 \)[/tex] and [tex]\( q = -4 \)[/tex].
- Sketched the curve based on roots and the behavior around them.
