Find an equation for the tangent line to the curve at the given point. Then sketch the curve and the tangent line together. \[ y=8 \sqrt{x},(4,16) \] An equation for the tangent line is (Type an equation.)

To find the equation of the tangent line to the curve \(y = 8\sqrt{x}\) at the point \((4, 16)\), we first need to calculate the derivative of the function, which gives us the slope of the tangent line. The derivative of \(y = 8\sqrt{x}\) is: \[ \frac{dy}{dx} = \frac{8}{2\sqrt{x}} = \frac{4}{\sqrt{x}}. \] Now, we substitute \(x = 4\) to find the slope at that point: \[ \frac{dy}{dx} \bigg|_{x=4} = \frac{4}{\sqrt{4}} = \frac{4}{2} = 2. \] We have the slope \(m = 2\) and the point \((4, 16)\). Using the point-slope form of a line, we can write the equation of the tangent line: \[ y - 16 = 2(x - 4). \] Simplifying, we get: \[ y = 2x + 8. \] So, the equation for the tangent line is: \[ y = 2x + 8. \] Now, for the graphical representation, you would sketch the curve \(y = 8\sqrt{x}\) which starts at the origin and curves upwards, along with the straight line \(y = 2x + 8\) that touches the curve at \(x = 4\). It’s comforting to know that derivatives help us understand how functions behave, much like finding the perfect route for a road! When sketching, remember that the curve will rise steeply as \(x\) increases, while the tangent line will continue infinitely straight, intersecting the curve only at that point.