Answer:
To find the slope of the tangent line to the curve \(y = -\sqrt{8x + 1}\) at the point where \(x = 3\), we can use the limit definition of the derivative. The derivative of \(y\) with respect to \(x\), denoted as \(y'\) or \(\frac{dy}{dx}\), gives us the slope of the tangent line.
Here's how you can find the derivative:
\[y = -\sqrt{8x + 1}\]
Now, let's find \(y'\) by taking the derivative with respect to \(x\):
\[y' = \frac{d}{dx}\left(-\sqrt{8x + 1}\right)\]
We'll use the chain rule for differentiation here, where if \(u = 8x + 1\), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).
1. Find \(\frac{dy}{du}\):
\[\frac{dy}{du} = \frac{d}{du}\left(-\sqrt{u}\right) = -\frac{1}{2\sqrt{u}}\]
2. Find \(\frac{du}{dx}\):
\[\frac{du}{dx} = \frac{d}{dx}\left(8x + 1\right) = 8\]
Now, multiply these two derivatives together:
\[y' = \left(-\frac{1}{2\sqrt{u}}\right) \cdot 8 = -\frac{4}{\sqrt{8x + 1}}\]
Now that we have the derivative, we can find the slope of the tangent line at \(x = 3\) by substituting \(x = 3\) into \(y'\):
\[y'(3) = -\frac{4}{\sqrt{8(3) + 1}} = -\frac{4}{\sqrt{25}} = -\frac{4}{5}\]
So, the slope of the tangent line at \(x = 3\) is \(-\frac{4}{5}\).
Now, let's find the equation of the tangent line using the point-slope form:
\[y - y_1 = m(x - x_1)\]
Where \(m\) is the slope we found (\(-\frac{4}{5}\)), and \((x_1, y_1)\) is the point of tangency (\(x = 3, y = -\sqrt{8(3) + 1}\)).
Plug in these values:
\[y - (-\sqrt{8(3) + 1}) = -\frac{4}{5}(x - 3)\]
Simplify:
\[y + \sqrt{25} = -\frac{4}{5}(x - 3)\]
Now, simplify further:
\[y + 5 = -\frac{4}{5}(x - 3)\]
To get the equation of the tangent line in slope-intercept form (\(y = mx + b\)), solve for \(y\):
\[y = -\frac{4}{5}(x - 3) - 5\]
Now, you have the equation of the tangent line to the curve \(y = -\sqrt{8x + 1}\) at the point where \(x = 3\):
\[y = -\frac{4}{5}(x - 3) - 5\]
Explanation: