https://file.aiquickdraw.com/mathbot/file/1727283256997yz9l3l04.jpg

Let's address each of the questions presented in the image.

### (i) Express \(2x^2 + 8x - 10\) in the form \(a(x+b)^2 + c\)

To complete the square:
1. Factor out the 2 from the first two terms: \(2(x^2 + 4x) - 10\).
2. Take half of the coefficient of \(x\) (which is 4), square it, and add and subtract this value inside the bracket: \(2(x^2 + 4x + 4 - 4) - 10\).
3. This gives \(2((x+2)^2 - 4) - 10\).
4. Simplify to get \(2(x+2)^2 - 8 - 10 = 2(x+2)^2 - 18\).

### (ii) For the curve \(y = 2x^2 + 8x - 10\), state the least value of \(y\) and the corresponding value of \(x\)

From the expression found in (i), we know the vertex form is \(2(x+2)^2 - 18\). The vertex of a parabola in this form represents the minimum point if \(a > 0\), which is the case here.

- The least value of \(y\) is at \(y = -18\).
- The corresponding value of \(x\) is at \(x = -2\).

### (iii) Find the set of values of \(x\) for which \(y \geq 14\)

Substituting \(y = 14\) into the vertex form and solving for \(x\):
\[ 2(x+2)^2 - 18 \geq 14 \]
\[ 2(x+2)^2 \geq 32 \]
\[ (x+2)^2 \geq 16 \]
\[ x+2 \geq 4 \text{ or } x+2 \leq -4 \]
\[ x \geq 2 \text{ or } x \leq -6 \]
So, \(x \in (-\infty, -6] \cup [2, \infty)\).

### (iv) Find the least value of \(k\) for which \(f\) is one-one for the domain \(x \geq k\)

Since the function has a minimum point at \(x = -2\), to ensure the function is one-to-one, \(k\) must start from the vertex onward to avoid repeating \(y\)-values.

- The least value of \(k\) is \(-2\).

### (v) Express \(f^{-1}(x)\) in terms of \(x\) in this case

The function, restricted to \(x \geq -2\), is given by \(f(x) = 2(x+2)^2 - 18\).
1. Solve for \(x\) in terms of \(y\): \(y = 2(x+2)^2 - 18\)
2. Rearrange: \(2(x+2)^2 = y + 18\)
3. \((x+2)^2 = \frac{y + 18}{2}\)
4. \(x + 2 = \pm\sqrt{\frac{y + 18}{2}}\) (choose positive since \(x \geq -2\))
5. \(x = -2 + \sqrt{\frac{y + 18}{2}}\)

Thus, the inverse function for \(x \geq -2\) is:
\[ f^{-1}(x) = -2 + \sqrt{\frac{x + 18}{2}} \]

Would you like further details on any of these solutions, or do you have any other questions?

Here are some possible follow-up questions you might consider:
1. How do we derive the vertex form of a parabola from its standard form?
2. Why does selecting \(k = -2\) ensure the function is one-to-one for \(x \geq k\)?
3. What is the significance of the vertex in a quadratic function's graph?
4. Can we determine the range of the function for \(x \geq -2\)?
5. How do transformations of the quadratic function affect its graph?

**Tip:** When working with quadratic functions, completing the square can provide valuable insights into the function's properties, such as its vertex and line of symmetry, which are useful for solving various application problems.

6. (i) Express 2x^2 + 8x - 10 in the form a(x + b)^2 + c. (ii) For the curve y = 2x^2 + 8x - 10, state the least value of y and the corresponding value of x. (iii) Find the set of values of x for which y ≥ 14. Given that f: x → 2x^2 + 8x - 10 for the domain x ≥ k, (iv) find the least value of k for which f is one-one, (v) express f^{-1}(x) in terms of x in this case.

This problem involves completing the square for the quadratic equation 2x^2 + 8x - 10, analyzing the vertex and least value of y, determining when y ≥ 14, and finding the inverse function of f(x). A detailed step-by-step solution includes key algebraic concepts, making it ideal for Grades 10-12 students.

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