Find a parabola with the equation y = ax^2 + bx + c that has slope 4 at x=1, slope -8 at x=-1, and passes through the point (1,8).

We are tasked with finding the equation of a parabola of the form \( y = ax^2 + bx + c \) that meets the following conditions:

1. The slope at \( x = 1 \) is 4.
2. The slope at \( x = -1 \) is -8.
3. The parabola passes through the point \( (1, 8) \).

### Step 1: Find the derivative (slope) of the equation
The derivative of \( y = ax^2 + bx + c \) gives the slope of the parabola at any point \( x \):
\[
\frac{dy}{dx} = 2ax + b.
\]

### Step 2: Use the given slope conditions
Using the fact that the slope is 4 at \( x = 1 \), we substitute into the derivative:
\[
2a(1) + b = 4 \quad \Rightarrow \quad 2a + b = 4. \tag{1}
\]
Next, using the fact that the slope is -8 at \( x = -1 \), we substitute into the derivative again:
\[
2a(-1) + b = -8 \quad \Rightarrow \quad -2a + b = -8. \tag{2}
\]

### Step 3: Solve for \( a \) and \( b \)
We now have a system of two equations:
\[
2a + b = 4 \tag{1}
\]
\[
-2a + b = -8 \tag{2}
\]
Subtract equation (2) from equation (1) to eliminate \( b \):
\[
(2a + b) - (-2a + b) = 4 - (-8) \quad \Rightarrow \quad 4a = 12 \quad \Rightarrow \quad a = 3.
\]

Substitute \( a = 3 \) into equation (1):
\[
2(3) + b = 4 \quad \Rightarrow \quad 6 + b = 4 \quad \Rightarrow \quad b = -2.
\]

### Step 4: Use the condition that the parabola passes through \( (1, 8) \)
We now have \( a = 3 \) and \( b = -2 \), so the equation of the parabola is \( y = 3x^2 - 2x + c \). To find \( c \), we use the fact that the parabola passes through the point \( (1, 8) \). Substituting \( x = 1 \) and \( y = 8 \) into the equation:
\[
8 = 3(1)^2 - 2(1) + c \quad \Rightarrow \quad 8 = 3 - 2 + c \quad \Rightarrow \quad 8 = 1 + c \quad \Rightarrow \quad c = 7.
\]

### Final Equation of the Parabola
Thus, the equation of the parabola is:
\[
y = 3x^2 - 2x + 7.
\]

Would you like further details or have any questions? Here are 5 related questions to explore:

1. How do you verify if the parabola meets the given slope conditions at specific points?
2. How does changing the slopes affect the coefficients of a quadratic equation?
3. Can a parabola have horizontal tangents at specific points? If so, how?
4. How can you find the vertex of the parabola from its equation?
5. What is the significance of the point \( (1, 8) \) in determining the equation?

**Tip**: The derivative of a quadratic function gives a linear equation, which makes it easy to calculate the slope at any given point!

This problem demonstrates how to find the equation of a parabola y = ax^2 + bx + c that meets specific slope conditions at given points and passes through a specified point. We solve for the coefficients using calculus and algebra, ideal for students in Grades 10-12.

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