Question

Use the limit definition to find an equation of the tangent line to the graph of \(f\) at the given point. Then verify your results by using a graphing utility to graph the function and its tangent line at the point. $$ f(x)=(x-1)^{2} ;(-2,9) $$

Short answer

The equation of the tangent line to the graph of the function \(f(x)=(x-1)^{2}\) at the point (-2,9) is \(y = -6x -3\). The results are verified by graphing the function and the tangent line.

Step-by-step solution

Find the derivative of the function

The derivative of a function at a particular point gives the slope of the tangent line at that point. The limit definition of the derivative is defined as \(f'(x)= \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}\). Let's find the derivative of \(f(x)=(x-1)^{2}\) using this definition. In this case, \(f(x+h)=(x+h-1)^2\), so \(f'(x) = \lim_{h\rightarrow 0} \frac{(x+h-1)^2-(x-1)^2}{h}\). After simplifying, we find that \(f'(x) = 2x - 2\).

Find the slope of the tangent line at the given point

Substitute the x-value of the given point into the derivative function to find the slope of the tangent line. The x-value of the given point (-2,9) is -2. Let’s substitute x = -2 into \(f'(x) = 2x - 2\). This gives us \(f'(-2) = 2*-2 - 2 = -6\). Therefore, the slope of the tangent line at the point (-2,9) is -6.

Find the equation of the tangent line

We can use the point-slope form of a line equation to find the equation of the tangent line. The point-slope form of a line is \(y - y1 = m*(x - x1)\), where m is the slope of the line, and (x1, y1) is the point that the line passes through. Let’s substitute m = -6, x1 = -2, and y1 = 9 into the equation. This gives us \(y - 9 = -6 * (x + 2)\). After simplifying, we get the equation of the tangent line as \( y = -6x - 3\).

Graph the function and the tangent line

To verify our results, use a graphing utility to graph the function \(y = (x-1)²\) and its tangent line \(y= -6x -3 \). The tangent line should touch the graph of the function only at the point (-2,9).