Let's go through each problem step-by-step, showing the full work for how they are solved.

### Problem 41: 
**Vertex:** \((1, -2)\), **Point:** \((-1, 14)\)

The equation of a parabola in vertex form is:
\[
y = a(x - h)^2 + k
\]
where \((h, k)\) is the vertex. Here, the vertex is \((1, -2)\), so we substitute \(h = 1\) and \(k = -2\) into the equation:
\[
y = a(x - 1)^2 - 2
\]
Now, use the point \((-1, 14)\) to find \(a\). Substituting \(x = -1\) and \(y = 14\) into the equation:
\[
14 = a(-1 - 1)^2 - 2
\]
Simplifying:
\[
14 = a(-2)^2 - 2
\]
\[
14 = 4a - 2
\]
Add 2 to both sides:
\[
16 = 4a
\]
Solve for \(a\):
\[
a = 4
\]
Thus, the equation of the parabola is:
\[
y = 4(x - 1)^2 - 2
\]

---

### Problem 42: 
**Vertex:** \((-4, -1)\), **Point:** \((-2, -4)\)

The equation of the parabola is:
\[
y = a(x - h)^2 + k
\]
Substitute \(h = -4\) and \(k = -1\):
\[
y = a(x + 4)^2 - 1
\]
Use the point \((-2, -4)\) to solve for \(a\):
\[
-4 = a(-2 + 4)^2 - 1
\]
Simplifying:
\[
-4 = a(2)^2 - 1
\]
\[
-4 = 4a - 1
\]
Add 1 to both sides:
\[
-3 = 4a
\]
Solve for \(a\):
\[
a = -\frac{3}{4}
\]
Thus, the equation of the parabola is:
\[
y = -\frac{3}{4}(x + 4)^2 - 1
\]

---

### Problem 43: 
**Vertex:** \(\left(\frac{1}{2}, 1\right)\), **Point:** \(\left(-2, \frac{21}{

This guide provides a detailed step-by-step solution to solve problems involving parabolas in vertex form. Learn how to use points and vertices to find the coefficient 'a' and derive the full equation. Suitable for Grades 9-11.

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