7) Find the Tangent line to \( y=(x+3)^{2}-2 \) at \( x=3 \)

To find the tangent line to the curve at a given point, we first need to determine the derivative of the function. The original function is \( y = (x+3)^2 - 2 \). The derivative, using the chain rule, is \( y' = 2(x+3) \). Evaluating the derivative at \( x = 3 \) gives us \( y'(3) = 2(3+3) = 12 \), which means the slope of the tangent line is 12. Next, we find the corresponding \( y \)-value at \( x = 3 \): \( y(3) = (3+3)^2 - 2 = 36 - 2 = 34 \). Now we have a point \( (3, 34) \) and a slope of 12. Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \), we plug in our values: \( y - 34 = 12(x - 3) \). This simplifies to \( y = 12x - 12 + 34 \), or \( y = 12x + 22 \). For someone studying calculus, working with derivatives can feel like riding a roller coaster: there are steep climbs and exhilarating drops! The first step is to ensure you've grasped the basic idea of derivatives as slopes of tangents. If you're unsure, try sketching the curve, the point, and the tangent line. It’s like connecting the dots but with a thrilling twist! Applying tangent lines in real-world scenarios is pretty nifty too! For instance, if you were to model the path of a thrown ball or the speed of a car at a specific moment, the tangent line would show you the instantaneous rate of change at that exact point. This is super useful in physics for calculating momentum, velocity, or even when designing roller coasters—it’s all about ensuring that the ride is thrilling but safe!