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To find the cubic function \( y = ax^3 + bx^2 + cx + d \) whose graph has horizontal tangents at the points \( (-2, 6) \) and \( (2, 0) \), we will follow these steps:

### Step 1: Use the condition for horizontal tangents.
Horizontal tangents mean that the derivative of the function is zero at those points. The derivative of the cubic function \( y = ax^3 + bx^2 + cx + d \) is:
\[
y' = 3ax^2 + 2bx + c
\]
For horizontal tangents at \( x = -2 \) and \( x = 2 \), we set the derivative equal to zero at those points:
\[
y'(-2) = 0 \quad \text{and} \quad y'(2) = 0
\]

Substitute \( x = -2 \) and \( x = 2 \) into the derivative:
\[
3a(-2)^2 + 2b(-2) + c = 0 \quad \Rightarrow \quad 12a - 4b + c = 0
\]
\[
3a(2)^2 + 2b(2) + c = 0 \quad \Rightarrow \quad 12a + 4b + c = 0
\]

Now we have two equations:
1. \( 12a - 4b + c = 0 \)
2. \( 12a + 4b + c = 0 \)

### Step 2: Use the given points to find additional equations.
We know the function passes through the points \( (-2, 6) \) and \( (2, 0) \), so substitute these points into the cubic function:

For \( (-2, 6) \):
\[
a(-2)^3 + b(-2)^2 + c(-2) + d = 6 \quad \Rightarrow \quad -8a + 4b - 2c + d = 6
\]
For \( (2, 0) \):
\[
a(2)^3 + b(2)^2 + c(2) + d = 0 \quad \Rightarrow \quad 8a + 4b + 2c + d = 0
\]

Now we have two more equations:
3. \( -8a + 4b - 2c + d = 6 \)
4. \( 8a + 4b + 2c + d = 0 \)

### Step 3: Solve the system of equations.
We now have four equations:
1. \( 12a - 4b + c = 0 \)
2. \( 12a + 4b + c = 0 \)
3. \( -8a + 4b - 2c + d = 6 \)
4. \( 8a + 4b + 2c + d = 0 \)

By solving this system of equations, we can find \( a \), \( b \), \( c \), and \( d \).

Let me proceed with solving these for the constants.The solution to the system of equations gives the following values for the constants:

\[
a = \frac{3}{16}, \quad b = 0, \quad c = -\frac{9}{4}, \quad d = 3
\]

Thus, the cubic function is:
\[
y = \frac{3}{16}x^3 - \frac{9}{4}x + 3
\]

Would you like further details on any step or additional explanation? Here are some follow-up questions that may interest you:

1. How do you solve systems of linear equations manually?
2. Why do horizontal tangents imply that the derivative is zero?
3. What are the key properties of cubic functions?
4. How does the second derivative help in analyzing concavity of the cubic function?
5. What is the geometric meaning of horizontal tangents in this context?

**Tip**: When working with derivatives, always check critical points (like where the derivative is zero) for insights into the behavior of a function.

Find a cubic function y = ax^3 + bx^2 + cx + d whose graph has horizontal tangents at the points (-2, 6) and (2, 0).

Find the cubic function y = ax^3 + bx^2 + cx + d with horizontal tangents at (-2, 6) and (2, 0). Learn how to solve for coefficients using derivatives and points on the curve, covering key calculus concepts. Suitable for Grades 10-12.

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