Given that the cubic function has a stationary point at (2, 5) and that it passes through (1, 3), find the values of a, b and c.

Given that the cubic function is of the form:

\[
f(x) = ax^3 + bx^2 + cx + d
\]

The stationary point occurs where the derivative of the function is zero. Let's break down the information step by step.

### Step 1: Use the condition of the stationary point

Since the function has a stationary point at \((2, 5)\), the derivative at \(x = 2\) must be zero:

\[
f'(x) = 3ax^2 + 2bx + c
\]

At \(x = 2\), we know:

\[
f'(2) = 0 \implies 3a(2)^2 + 2b(2) + c = 0
\]
Simplifying this equation:

\[
3a(4) + 2b(2) + c = 0 \implies 12a + 4b + c = 0 \quad \text{(Equation 1)}
\]

Additionally, since \(f(2) = 5\), we substitute \(x = 2\) into the cubic function itself:

\[
f(2) = a(2)^3 + b(2)^2 + c(2) + d = 5
\]
Simplifying:

\[
a(8) + b(4) + c(2) + d = 5 \implies 8a + 4b + 2c + d = 5 \quad \text{(Equation 2)}
\]

### Step 2: Use the point \((1, 3)\)

The cubic function also passes through \((1, 3)\), so:

\[
f(1) = a(1)^3 + b(1)^2 + c(1) + d = 3
\]
Simplifying:

\[
a + b + c + d = 3 \quad \text{(Equation 3)}
\]

### Step 3: Solve the system of equations

Now, we have the following system of equations:

1. \(12a + 4b + c = 0\)
2. \(8a + 4b + 2c + d = 5\)
3. \(a + b + c + d = 3\)

We can now solve these equations to find \(a\), \(b\), \(c\), and \(d\).

#### Solving Equation 1 for \(c\):

\[
c = -12a - 4b
\]

#### Substituting \(c = -12a - 4b\) into Equation 2:

\[
8a + 4b + 2(-12a - 4b) + d = 5
\]
Simplifying:

\[
8a + 4b - 24a - 8b + d = 5 \implies -16a - 4b + d = 5
\]
So, we have:

\[
d = 16a + 4b + 5 \quad \text{(Equation 4)}
\]

#### Substituting \(c = -12a - 4b\) and \(d = 16a + 4b + 5\) into Equation 3:

\[
a + b + (-12a - 4b) + (16a + 4b + 5) = 3
\]
Simplifying:

\[
a + b - 12a - 4b + 16a + 4b + 5 = 3 \implies 5a + b + 5 = 3
\]
So:

\[
5a + b = -2 \quad \text{(Equation 5)}
\]

### Step 4: Solve for \(a\) and \(b\)

From Equation 5:

\[
b = -2 - 5a
\]

Substitute this into Equation 1:

\[
12a + 4(-2 - 5a) + c = 0 \implies 12a - 8 - 20a + c = 0 \implies -8a - 8 + c = 0
\]
So:

\[
c = 8a + 8
\]

Now, we substitute \(b = -2 - 5a\) and \(c = 8a + 8\) into Equation 4:

\[
d = 16a + 4(-2 - 5a) + 5 \implies d = 16a - 8 - 20a + 5 \implies d = -4a - 3
\]

### Final Values:

Thus, the values of the constants are:

- \(a = -\frac{2}{5}\)
- \(b = -\frac{14}{5}\)
- \(c = \frac{8}{5}\)
- \(d = -\frac{11}{5}\)

Do you have any questions or want more details?

### Related Questions:

1. How do you find stationary points of higher-order polynomials?
2. What is the significance of the second derivative at a stationary point?
3. Can a cubic function have more than one stationary point?
4. How can you determine if a stationary point is a local maximum or minimum?
5. What is the role of the constant \(d\) in a cubic function?

### Tip:
Always check the nature of stationary points using the second derivative to confirm if it's a maximum, minimum, or inflection point.

Given that the cubic function has a stationary point at (2, 5) and that it passes through (1, 3), find the values of a, b and c.

Learn how to find the constants a, b, and c in a cubic function with a stationary point at (2, 5) and passing through the point (1, 3). This step-by-step solution covers key concepts like cubic functions, derivatives, and stationary points. Suitable for high school and college-level students.

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