Given a cubic function $f(x) = ax^3 + bx^2 + cx + d$, we need to find the coefficients $a, b, c,$ and $d$. We are given that the function has a relative maximum at $y=4$ when $x=1$ and that the graph touches the x-axis at $x=3$. In the second part, we need to analyze and graph $f(x) = x^3 - 6x^2 + 9x$.

1. Problem Description

Given a cubic function $f(x) = ax^3 + bx^2 + cx + d$, we need to find the coefficients $a, b, c,$ and $d$. We are given that the function has a relative maximum at $y=4$ when $x=1$ and that the graph touches the x-axis at $x=3$. In the second part, we need to analyze and graph $f(x) = x^3 - 6x^2 + 9x$.

2. Solution Steps

Part a:

We have $f(x) = ax^3 + bx^2 + cx + d$.

Since the graph touches the x-axis at $x=3$, we know that $f(3) = 0$. Thus,

$27a + 9b + 3c + d = 0$. (1)

Since there is a relative maximum at $x=1$ and $y=4$, we have $f(1) = 4$. Thus,

$a + b + c + d = 4$. (2)

At the relative maximum, the derivative is zero, so $f'(1) = 0$.

First, we find the derivative: $f'(x) = 3ax^2 + 2bx + c$.

So, $3a + 2b + c = 0$. (3)

Since the graph touches the x-axis at $x=3$, $x=3$ is a repeated root, meaning the function can be written as $f(x) = a(x-3)^2 (x-r)$ for some value $r$. However, the given information says the graph *touches* the x-axis at $x=3$, which means the curve does not cross the axis at $x=3$. This implies that $x=3$ is a repeated root.

Let's analyze the function for $f(x) = x^3 - 6x^2 + 9x$. This is the function in part b. The function given in part a is different. Let us continue with part b first.

Part b: $f(x) = x^3 - 6x^2 + 9x$

First, we find the derivative: $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$.

Setting $f'(x) = 0$, we find critical points at $x=1$ and $x=3$.

The second derivative is $f''(x) = 6x - 12$.

$f''(1) = 6(1) - 12 = -6 < 0$, so we have a local maximum at $x=1$.

$f''(3) = 6(3) - 12 = 6 > 0$, so we have a local minimum at $x=3$.

The values of the function at these points are:

$f(1) = 1^3 - 6(1)^2 + 9(1) = 1 - 6 + 9 = 4$.

$f(3) = 3^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0$.

So, the function has a local maximum at $(1, 4)$ and a local minimum at $(3, 0)$.

Since the function touches the x-axis at x=3, its derivative is also 0 at x=

3. $f(x) = x(x^2 - 6x + 9) = x(x-3)^2$.

Since $f(x) = x^3 - 6x^2 + 9x = ax^3 + bx^2 + cx + d$,

we have $a=1, b=-6, c=9, d=0$. This also satisfies $f(1)=4$, as needed.

Since we obtained $a=1$, $b=-6$, $c=9$, and $d=0$ by analyzing the given function in the second part of the question, there seems to be an implicit assumption that this function is supposed to be used for the first part as well, which is confusingly phrased in the problem statement.

So let us assume $a=1, b=-6, c=9, d=0$.

3. Final Answer

Part a:

Assuming the function from part b is to be used for part a, the coefficients are:

$a = 1$

$b = -6$

$c = 9$

$d = 0$

Part b:

The function $f(x) = x^3 - 6x^2 + 9x$ has a local maximum at $(1, 4)$ and a local minimum at $(3, 0)$. It touches the x-axis at $x=3$.