find the slope of the tangent line to the curve x^2+2xy-y^2+x=39 at x=5 with a negative y intercept

We are asked to find the slope of the tangent line to the curve \(x^2 + 2xy - y^2 + x = 39\) at \(x = 5\), where the tangent line has a negative y-intercept.

### Step 1: Implicit Differentiation
To find the slope of the tangent line, we need the derivative \(\frac{dy}{dx}\) using implicit differentiation. Differentiate both sides of the equation with respect to \(x\).

\[
\frac{d}{dx} \left( x^2 + 2xy - y^2 + x \right) = \frac{d}{dx}(39)
\]

Since the derivative of a constant (39) is zero, we focus on differentiating the left side:

1. \( \frac{d}{dx}(x^2) = 2x \)
2. \( \frac{d}{dx}(2xy) = 2y + 2x \frac{dy}{dx} \) (using the product rule)
3. \( \frac{d}{dx}(-y^2) = -2y \frac{dy}{dx} \)
4. \( \frac{d}{dx}(x) = 1 \)

Thus, the differentiated equation becomes:

\[
2x + 2y + 2x \frac{dy}{dx} - 2y \frac{dy}{dx} + 1 = 0
\]

### Step 2: Solve for \(\frac{dy}{dx}\)
Now, isolate \(\frac{dy}{dx}\):

\[
2x + 2y + 1 + (2x - 2y) \frac{dy}{dx} = 0
\]

Move all terms not involving \(\frac{dy}{dx}\) to the right-hand side:

\[
(2x - 2y) \frac{dy}{dx} = -(2x + 2y + 1)
\]

Solve for \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} = \frac{-(2x + 2y + 1)}{2x - 2y}
\]

### Step 3: Evaluate the Derivative at \(x = 5\)
We need the value of \(y\) corresponding to \(x = 5\). Substitute \(x = 5\) into the original equation:

\[
(5)^2 + 2(5)(y) - y^2 + 5 = 39
\]

Simplify:

\[
25 + 10y - y^2 + 5 = 39
\]

\[
30 + 10y - y^2 = 39
\]

Rearrange into a quadratic equation:

\[
y^2 - 10y + 9 = 0
\]

Solve this quadratic equation using the quadratic formula:

\[
y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(9)}}{2(1)}
\]

\[
y = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2}
\]

Thus, the solutions are:

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

### Step 4: Choose the Correct \(y\)-value
The problem specifies that the tangent line has a negative y-intercept. Therefore, we choose \(y = 9\), because the slope of the tangent line for \(y = 1\) results in a positive y-intercept (which would contradict the given condition).

### Step 5: Find the Slope
Now substitute \(x = 5\) and \(y = 9\) into the derivative formula:

\[
\frac{dy}{dx} = \frac{-(2(5) + 2(9) + 1)}{2(5) - 2(9)} = \frac{-(10 + 18 + 1)}{10 - 18} = \frac{-(29)}{-8} = \frac{29}{8}
\]

Thus, the slope of the tangent line at \(x = 5\) is \(\frac{29}{8}\).

Would you like a detailed explanation on any step? Here are 5 related questions:

1. How do we apply implicit differentiation for complex equations?
2. Can you explain how the quadratic formula was derived in detail?
3. What does the slope of a tangent line represent geometrically?
4. What are the conditions for determining a negative y-intercept?
5. How can we interpret the results if the y-values had different signs?

**Tip:** When solving quadratic equations, always carefully check the signs and potential extraneous solutions.

Find the slope of the tangent line to the curve x^2 + 2xy - y^2 + x = 39 at x = 5 with a negative y-intercept.

Learn how to find the slope of the tangent line to the curve x^2 + 2xy - y^2 + x = 39 at x = 5 with a detailed step-by-step solution. This problem covers implicit differentiation, solving quadratic equations, and analyzing the slope for a negative y-intercept. Suitable for students in Grades 10-12.

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