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Let's go through the problem step by step.

### 1. Given the function defined as:

\[
f(x) = x^3 - \frac{3}{2} x^2 - 6x + 10
\]

#### a) Find the point(s) of **horizontal tangency** of \( f(x) \).

- To find horizontal tangency, we need to find where the slope of the tangent line is zero. This occurs where the derivative of the function equals zero.

First, find the derivative of \( f(x) \):

\[
f'(x) = 3x^2 - 3x - 6
\]

Now, set \( f'(x) = 0 \) to find the critical points:

\[
3x^2 - 3x - 6 = 0
\]

Simplifying:

\[
x^2 - x - 2 = 0
\]

This can be factored:

\[
(x - 2)(x + 1) = 0
\]

So, the solutions are \( x = 2 \) and \( x = -1 \).

Now, we find the corresponding \( y \)-values by plugging these \( x \)-values into the original function \( f(x) \).

- For \( x = 2 \):

\[
f(2) = 2^3 - \frac{3}{2}(2)^2 - 6(2) + 10 = 8 - 6 - 12 + 10 = 0
\]

- For \( x = -1 \):

\[
f(-1) = (-1)^3 - \frac{3}{2}(-1)^2 - 6(-1) + 10 = -1 - \frac{3}{2} + 6 + 10 = \frac{19}{2}
\]

Thus, the points of horizontal tangency are \( (2, 0) \) and \( (-1, \frac{19}{2}) \).

#### b) Write an equation of the line **normal** (perpendicular) to the graph of \( f \) at \( x = 0 \).

To find the equation of the normal line, we first need the slope of the tangent line at \( x = 0 \). We do this by calculating \( f'(0) \):

\[
f'(0) = 3(0)^2 - 3(0) - 6 = -6
\]

The slope of the tangent line at \( x = 0 \) is \( -6 \), so the slope of the normal line will be the negative reciprocal of \( -6 \), which is \( \frac{1}{6} \).

Next, find the point on the graph where \( x = 0 \):

\[
f(0) = 0^3 - \frac{3}{2}(0)^2 - 6(0) + 10 = 10
\]

So, the point is \( (0, 10) \).

Now, use the point-slope form to write the equation of the normal line:

\[
y - 10 = \frac{1}{6}(x - 0)
\]

Simplifying:

\[
y = \frac{1}{6}x + 10
\]

#### c) Find the \( x \)- and \( y \)-coordinates of all points on the graph of \( f \) where the line tangent to the graph is **parallel to the \( x \)-axis** (i.e., where the slope is 0).

This was already found in part (a). The points of horizontal tangency occur where the slope is 0, which are at:

\[
(2, 0) \text{ and } (-1, \frac{19}{2})
\]

---

Would you like a more detailed explanation on any step or further clarifications?

Here are some related questions to deepen your understanding:

1. How do you determine whether a critical point corresponds to a minimum, maximum, or neither?
2. What is the geometric interpretation of a normal line to a curve?
3. Why does the slope of the normal line become the negative reciprocal of the tangent slope?
4. Can a function have more than one horizontal tangent? Under what conditions?
5. How does factoring help in solving quadratic equations derived from derivatives?

**Tip:** When solving for points of horizontal tangency, remember that you are looking for where the derivative equals zero, which gives you the critical points. Then evaluate the original function to find the corresponding \( y \)-coordinates.

Given the function defined as f(x) = x^3 - (3/2) x^2 - 6x + 10, find points of horizontal tangency, the equation of the normal line at x = 0, and coordinates where the tangent is parallel to the x-axis.

In this AP Calculus problem, learn how to find points of horizontal tangency, the equation of the normal line at x = 0, and coordinates where the tangent line is parallel to the x-axis for the function f(x) = x^3 - (3/2)x^2 - 6x + 10.

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