Let's work through each question step-by-step.

Part (c)

The goal is to rewrite $f(x) = 2x^2 + 12x + 14$ in the form $f(x) = a(x + h)^2 + k$, which represents the vertex form of a quadratic function. This will allow us to determine the turning point of $f(x)$.

1. Rewrite $f(x)$: $f(x) = 2x^2 + 12x + 14$ Factor out the 2 from the $x^2$ and $x$ terms: $f(x) = 2(x^2 + 6x) + 14$

2. Complete the square: Inside the parentheses, take half of 6 (which is 3), square it to get 9, and add and subtract this value: $f(x) = 2(x^2 + 6x + 9 - 9) + 14$ Simplify within the parentheses: $f(x) = 2((x + 3)^2 - 9) + 14$ Distribute the 2: $f(x) = 2(x + 3)^2 - 18 + 14$ Combine constants: $f(x) = 2(x + 3)^2 - 4$ So, $f(x) = a(x + h)^2 + k$ with $a = 2$, $h = -3$, and $k = -4$.

3. Turning point: The turning point (vertex) of $f(x)$ is $(-h, k) = (-(-3), -4) = (3, -4)$.

Part (d)

We need to solve the equation: $\left| \frac{x - 2}{x + 3} \right| = 4$

To solve this, consider the two cases for the absolute value expression.

1. Case 1: $\frac{x - 2}{x + 3} = 4$ Multiply both sides by $x + 3$ (assuming $x \neq -3$): $x - 2 = 4(x + 3)$ Expand and simplify: $x - 2 = 4x + 12$ $-3x = 14$ $x = -\frac{14}{3}$

2. Case 2: $\frac{x - 2}{x + 3} = -4$ Multiply both sides by $x + 3$ (assuming $x \neq -3$): $x - 2 = -4(x + 3)$ Expand and simplify: $x - 2 = -4x - 12$ $5x = -10$ $x = -2$

So, the solutions for $x$ are $x = -\frac{14}{3}$ and $x = -2$.

Summary

• Part (c): The vertex form of $f(x) = 2x^2 + 12x + 14$ is $f(x) = 2(x + 3)^2 - 4$, with a turning point at $(3, -4)$.
• Part (d): The values of $x$ that satisfy the equation are $x = -\frac{14}{3}$ and $x = -2$.

Would you like further details on any part of the solution, or do you have any questions?

Related Questions

1. How can we derive the vertex form of a quadratic equation by completing the square?
2. Why does completing the square help in finding the turning point of a quadratic function?
3. What is the geometric interpretation of the vertex form of a quadratic function?
4. How does the absolute value function affect the solutions to an equation?
5. What alternative methods can solve quadratic equations involving absolute values?

Tip

To solve absolute value equations, always consider both positive and negative scenarios within the expression.