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When it comes to finding the equation of the tangent line to a curve, the first essential step is to calculate the derivative of the curve's function. The derivative tells us the rate at which the function's value changes with respect to changes in the independent variable. Essentially, it gives us the slope of the tangent at any point on the curve. For the function given as: \[ y = 4 - x^2 \] Here's how we calculate its derivative:

• The derivative of a constant, like 4, is 0.
• The derivative of \(-x^2\) is \(-2x\), by applying the power rule, which states: bring down the exponent and multiply it by the function's coefficient, then reduce the exponent by 1.

So, the derivative of the curve function is:\[ y' = -2x \]This equation represents the slope of the tangent line to the curve at any given point \(x\).

To find the slope of the tangent at a specific point on the curve, we evaluate the derivative at that point. This process requires substituting the x-value of the point into the derivative.In our case, the point given is \((-1, 3)\). By substituting \(x = -1\) into the derivative:\[ y' = -2(-1) = 2 \]The slope of the tangent line, \(m\), at the point \((-1, 3)\) is 2. This tells us that at this point, the tangent line is rising at a rate of 2 units up for every unit it moves to the right. Understanding the slope is crucial as it determines the angle and steepness of the tangent line.

Once we have the slope of the tangent line, we can find the equation of that line using the point-slope form. This is a simple and efficient way to form an equation of a line when a point on the line and the slope are known.The point-slope formula is:\[ y - y_1 = m(x - x_1) \]where \((x_1, y_1)\) is a given point on the line and \(m\) is the slope.Given our point \((-1, 3)\) and slope \(m = 2\), we substitute these values into the formula:\[ y - 3 = 2(x + 1) \]After simplifying, we find the equation in slope-intercept form (\(y = mx + b\)), which is easier to graph:\[ y = 2x + 5 \]This equation describes the tangent line precisely as touching the curve only at \((-1, 3)\). Using the point-slope form to derive the equation is a fundamental technique in algebra and calculus.