To find the tangents to the curve $f(x) = (x - 1)(x - 2)(x - 3)$ at all $x$-intercepts, let's go through the steps in detail.

Step 1: Find the $x$-intercepts

The $x$-intercepts occur where $f(x) = 0$. Therefore, we set $(x - 1)(x - 2)(x - 3) = 0.$ This gives the roots:

1. $x = 1$
2. $x = 2$
3. $x = 3$

So, the $x$-intercepts are at $x = 1$, $x = 2$, and $x = 3$.

Step 2: Differentiate $f(x)$

To find the tangent lines, we need the derivative of $f(x)$, $f'(x)$, which gives us the slope of the tangent line at any point $x$.

The function $f(x) = (x - 1)(x - 2)(x - 3)$ is a product of three linear terms. To differentiate it, we can use the product rule. However, to simplify, let's expand $f(x)$ first.

Expanding $f(x)$: $f(x) = (x - 1)(x - 2)(x - 3) = (x^2 - 3x + 2)(x - 3).$ Now expand further: $f(x) = x^3 - 3x^2 + 2x - 3x^2 + 9x - 6 = x^3 - 6x^2 + 11x - 6.$

Now, differentiate $f(x)$: $f'(x) = 3x^2 - 12x + 11.$

Step 3: Find the slope of the tangent at each $x$-intercept

Now we substitute each $x$-intercept into $f'(x)$ to find the slope of the tangent line at that point.

1. At $x = 1$: $f'(1) = 3(1)^2 - 12(1) + 11 = 3 - 12 + 11 = 2.$ So, the slope of the tangent at $x = 1$ is $2$.

2. At $x = 2$: $f'(2) = 3(2)^2 - 12(2) + 11 = 3 \cdot 4 - 24 + 11 = 12 - 24 + 11 = -1.$ So, the slope of the tangent at $x = 2$ is $-1$.

3. At $x = 3$: $f'(3) = 3(3)^2 - 12(3) + 11 = 3 \cdot 9 - 36 + 11 = 27 - 36 + 11 = 2.$ So, the slope of the tangent at $x = 3$ is $2$.

Step 4: Find the equation of the tangent lines

The equation of a tangent line at a point $(a, f(a))$ with slope $m$ is given by: $y - f(a) = m(x - a).$

Now, let's apply this for each $x$-intercept.

1. At $x = 1$:

   • $f(1) = (1 - 1)(1 - 2)(1 - 3) = 0$.
   • Slope $m = 2$.

   The equation of the tangent is: $y - 0 = 2(x - 1) \Rightarrow y = 2x - 2.$

2. At $x = 2$:

   • $f(2) = (2 - 1)(2 - 2)(2 - 3) = 0$.
   • Slope $m = -1$.

   The equation of the tangent is: $y - 0 = -1(x - 2) \Rightarrow y = -x + 2.$

3. At $x = 3$:

   • $f(3) = (3 - 1)(3 - 2)(3 - 3) = 0$.
   • Slope $m = 2$.

   The equation of the tangent is: $y - 0 = 2(x - 3) \Rightarrow y = 2x - 6.$

Final Answer

The equations of the tangent lines to the curve at each $x$-intercept are:

1. At $x = 1$: $y = 2x - 2$
2. At $x = 2$: $y = -x + 2$
3. At $x = 3$: $y = 2x - 6$

Would you like further clarification or details on any part of this solution?

Here are five additional questions to explore this topic further:

1. How would the solution change if the function had four roots instead of three?
2. What is the significance of the slope being zero at certain points for the tangent lines?
3. How can we confirm that these points are indeed $x$-intercepts algebraically?
4. What would be the effect of adding a constant to $f(x)$ on the $x$-intercepts?
5. How does the product rule for derivatives apply when differentiating polynomials?

Tip: Expanding polynomials before differentiating can simplify the process, especially when dealing with products of multiple terms.