Determine the area (In -x -y) of the figure where y=8x and y=x^2-9. What is the shaded region in terms of y?

We are asked to find the area of the region enclosed between the curves \( y = 8x \) and \( y = x^2 - 9 \) in the negative \(x\)- and \(y\)-axis region. Let's break this down step by step.

### 1. **Determine points of intersection**:
To find the bounds for the area, we need to determine where the two curves intersect by solving \( 8x = x^2 - 9 \).

\[
8x = x^2 - 9
\]

Rearranging this into a standard quadratic form:

\[
x^2 - 8x - 9 = 0
\]

We can solve this quadratic equation using the quadratic formula:

\[
x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-9)}}{2(1)}
\]

\[
x = \frac{8 \pm \sqrt{64 + 36}}{2} = \frac{8 \pm \sqrt{100}}{2} = \frac{8 \pm 10}{2}
\]

Thus, the two solutions are:

\[
x = \frac{8 + 10}{2} = 9 \quad \text{and} \quad x = \frac{8 - 10}{2} = -1
\]

So, the points of intersection are at \( x = 9 \) and \( x = -1 \). However, since we are interested in the region on the negative \(x\)-axis, we will consider \( x = -1 \).

### 2. **Set up the integral for the area**:
The area between two curves is given by the integral of the difference between the top curve and the bottom curve over the interval. Here, \( y = 8x \) is the top curve, and \( y = x^2 - 9 \) is the bottom curve.

We are looking at the region where \( x \) goes from \(-1\) to \(0\) (since the area must be in the negative \(x\)-region).

The area \(A\) is:

\[
A = \int_{-1}^{0} \left(8x - (x^2 - 9)\right) \, dx
\]

Simplifying the integrand:

\[
A = \int_{-1}^{0} \left(8x - x^2 + 9\right) \, dx
\]

### 3. **Evaluate the integral**:

We can now integrate each term:

\[
\int_{-1}^{0} 8x \, dx = 4x^2 \Big|_{-1}^{0} = 4(0^2) - 4(-1^2) = 0 - 4 = -4
\]

\[
\int_{-1}^{0} x^2 \, dx = \frac{x^3}{3} \Big|_{-1}^{0} = \frac{0^3}{3} - \frac{(-1)^3}{3} = 0 + \frac{1}{3} = \frac{1}{3}
\]

\[
\int_{-1}^{0} 9 \, dx = 9x \Big|_{-1}^{0} = 9(0) - 9(-1) = 0 + 9 = 9
\]

Now, summing these results:

\[
A = (-4) - \left(\frac{1}{3}\right) + 9 = 5 - \frac{1}{3} = \frac{15}{3} - \frac{1}{3} = \frac{14}{3}
\]

### Final Answer:
The area of the region between the curves \( y = 8x \) and \( y = x^2 - 9 \) in the negative \(x\)- and \(y\)-axes is \( \frac{14}{3} \) square units.

Would you like further details or explanations?

Here are 5 related questions to consider:
1. How would the area change if the region was in the positive \(x\)-axis?
2. How do we interpret the negative area result if we don't restrict the \(x\)-region?
3. What happens if we translate one of the curves vertically?
4. How can we find the volume if the region is rotated around the \(x\)-axis?
5. Can we use numerical methods to approximate the area if exact integration is difficult?

**Tip:** Always sketch the curves to visually confirm the area you're integrating when solving such problems.

Discover how to find the area between the curves y=8x and y=x^2-9 in the negative x- and y-axis. This detailed step-by-step solution uses algebra and integration, solving the quadratic equation and calculating the area between two curves. Suitable for students in Grades 10-12.

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