https://file.aiquickdraw.com/mathbot/file/1730156741979gslh4npe.jpg

Let's break down and solve the problem shown in the image.

### Part (a): Find the coordinates of \(A\) and \(B\)
The two curves are:
- Quadratic curve: \( y = x^2 - 5x + 9 \)
- Line: \( y = 5 \)

We need to find the intersection points by setting the two equations equal to each other.

\[
x^2 - 5x + 9 = 5
\]

Simplify:
\[
x^2 - 5x + 4 = 0
\]

Now, solve this quadratic equation using the **quadratic formula**:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 1\), \(b = -5\), and \(c = 4\).

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)} = \frac{5 \pm \sqrt{25 - 16}}{2}
\]
\[
x = \frac{5 \pm \sqrt{9}}{2} = \frac{5 \pm 3}{2}
\]

This gives two solutions:
\[
x = \frac{5 + 3}{2} = 4 \quad \text{and} \quad x = \frac{5 - 3}{2} = 1
\]

Thus, the intersection points are:
- At \(x = 1\), \(y = 5\), giving point \(A(1, 5)\).
- At \(x = 4\), \(y = 5\), giving point \(B(4, 5)\).

### Part (b): Calculate the exact area of the finite region
We need to compute the area between the quadratic curve \(y = x^2 - 5x + 9\) and the line \(y = 5\) over the interval \(x = 1\) to \(x = 4\).

The area between two curves is given by:
\[
\text{Area} = \int_{x_1}^{x_2} \left( \text{Upper function} - \text{Lower function} \right) \, dx
\]

Here, the upper function is \(y = 5\), and the lower function is \(y = x^2 - 5x + 9\). So the integral becomes:
\[
\text{Area} = \int_1^4 \left( 5 - (x^2 - 5x + 9) \right) \, dx
\]
Simplify the integrand:
\[
5 - (x^2 - 5x + 9) = -x^2 + 5x - 4
\]
Now, integrate:
\[
\int_1^4 (-x^2 + 5x - 4) \, dx
\]

Step-by-step integration:
\[
\int (-x^2) \, dx = -\frac{x^3}{3}, \quad \int 5x \, dx = \frac{5x^2}{2}, \quad \int -4 \, dx = -4x
\]

Thus:
\[
\left[ -\frac{x^3}{3} + \frac{5x^2}{2} - 4x \right]_1^4
\]

Now, substitute the limits \(x = 4\) and \(x = 1\).

For \(x = 4\):
\[
-\frac{4^3}{3} + \frac{5(4^2)}{2} - 4(4) = -\frac{64}{3} + \frac{80}{2} - 16 = -\frac{64}{3} + 40 - 16 = -\frac{64}{3} + 24
\]

For \(x = 1\):
\[
-\frac{1^3}{3} + \frac{5(1^2)}{2} - 4(1) = -\frac{1}{3} + \frac{5}{2} - 4 = -\frac{1}{3} + 2.5 - 4 = -\frac{1}{3} - 1.5 = -\frac{5}{3}
\]

Now, subtract the two results:
\[
\left( -\frac{64}{3} + 24 \right) - \left( -\frac{5}{3} \right) = -\frac{64}{3} + 24 + \frac{5}{3}
\]
\[
= -\frac{64 - 5}{3} + 24 = -\frac{59}{3} + 24
\]
\[
= -\frac{59}{3} + \frac{72}{3} = \frac{72 - 59}{3} = \frac{13}{3}
\]

Thus, the exact area of the shaded region is:
\[
\boxed{\frac{13}{3}} \, \text{square units}.
\]

---

### Summary of Results:
- **(a)** The coordinates of \(A\) and \(B\) are \(A(1, 5)\) and \(B(4, 5)\).
- **(b)** The exact area of the shaded region is \(\frac{13}{3}\) square units.

---

Do you need further details or have any questions?

Here are five related questions:
1. How do we determine which function is the upper one in an area problem?
2. Can the quadratic formula always solve any quadratic equation?
3. How do we evaluate definite integrals over specific limits?
4. What does it mean if the area between two curves is negative?
5. How would the solution change if the line was \(y = 6\) instead?

**Tip:** When finding areas between curves, always double-check the limits of integration to avoid common errors.

The figure above shows a quadratic curve and a straight line with respective equations y = x^2 - 5x + 9 and y = 5. The points A and B are the points of intersection between the straight line and the quadratic curve. (a) Find the coordinates of A and B. (b) Calculate the exact area of the finite region bounded by the quadratic curve and the straight line, shown shaded in the above figure.

Learn how to find the intersection points and calculate the area of the region between a quadratic curve y = x^2 - 5x + 9 and a line y = 5. This problem involves solving a quadratic equation and computing a definite integral. Suitable for Grades 10-12.

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