Let's solve the questions one by one from the image.

Part (c)

You are asked to rewrite the expression: $f(x) = 2x^2 + 12x + 14$ in the form $f(x) = a(x + h)^2 + k$, where $a$, $h$, and $k$ are real numbers.

This requires us to complete the square for the expression $f(x)$.

1. Start with $f(x) = 2x^2 + 12x + 14$.
2. Factor out the coefficient of $x^2$, which is 2: $f(x) = 2(x^2 + 6x) + 14$
3. Complete the square for the expression inside the parentheses:
   • Take half of 6 (the coefficient of $x$), square it, and add/subtract it inside the parentheses.
   • Half of 6 is 3, and $3^2 = 9$.
4. Rewrite $x^2 + 6x$ as: $x^2 + 6x = (x + 3)^2 - 9$
5. Substitute back: $f(x) = 2((x + 3)^2 - 9) + 14$
6. Distribute the 2 and simplify: $f(x) = 2(x + 3)^2 - 18 + 14 = 2(x + 3)^2 - 4$ So, in the form $f(x) = a(x + h)^2 + k$, we have:

• $a = 2$
• $h = -3$
• $k = -4$

Thus, the rewritten form is: $f(x) = 2(x + 3)^2 - 4$

Turning Point of $f(x)$

The turning point of $f(x) = 2(x + 3)^2 - 4$ is given by the point $(-h, k)$. Here, $h = -3$ and $k = -4$, so the turning point is: $(-3, -4)$

Part (d)

You are asked to solve for $x$ in the equation: $\left| \frac{x - 2}{x + 3} \right| = 4$

This absolute value equation can be split into two cases:

Case 1: $\frac{x - 2}{x + 3} = 4$

1. Multiply both sides by $x + 3$: $x - 2 = 4(x + 3)$
2. Expand and simplify: $x - 2 = 4x + 12$
3. Move $x$ terms to one side: $-2 - 12 = 4x - x$ $-14 = 3x$ $x = -\frac{14}{3}$

Case 2: $\frac{x - 2}{x + 3} = -4$

1. Multiply both sides by $x + 3$: $x - 2 = -4(x + 3)$
2. Expand and simplify: $x - 2 = -4x - 12$
3. Move $x$ terms to one side: $x + 4x = -12 + 2$ $5x = -10$ $x = -2$

Solution

The values of $x$ that satisfy the equation are: $x = -\frac{14}{3} \quad \text{and} \quad x = -2$

Would you like further details on any step?

Related Questions

1. How do you verify the turning point of a quadratic function?
2. What are other methods for solving absolute value equations?
3. How is completing the square useful for graphing parabolas?
4. Can completing the square help in finding maximum or minimum values?
5. What are some real-life applications of quadratic functions?

Tip

When solving absolute value equations, always remember to consider both positive and negative cases.