Question

a. Find an equation of the line tangent to the given curve at a. b. Use a graphing utility to graph the curve and the tangent line on the same set of axes. $$y=x^{3}-4 x^{2}+2 x-1 ; a=2$$

Short answer

Answer: The equation of the tangent line is \(y + 5 = -2(x - 2)\).

Step-by-step solution

Find the derivative of the function

To find the derivative of the function \(y = x^3 - 4x^2 + 2x - 1\), use the power rule for each of the terms. The derivative, denoted as \(y'\) or \(\frac{dy}{dx}\), is given by: $$y' = \frac{d}{dx}(x^3 - 4x^2 + 2x - 1) = 3x^2 - 8x + 2$$

Evaluate the derivative at \(a = 2\)

Now we need to find the slope of the tangent line at \(a = 2\). Plug \(2\) into the derivative \(y'\): $$ y'(2)=3(2)^2-8(2)+2=12-16+2=-2 $$ The slope of the tangent line at \(a = 2\) is -2.

Find the equation of the tangent line

Now we will find the equation of the tangent line to the curve at \(a = 2\). First, find the value of \(y\) when \(x = 2\): $$ y(2)=2^3-4(2)^2+2(2)-1=8-16+4-1=-5 $$ The coordinate point on the curve at \(a = 2\) is \((2,-5)\). Using the slope of the tangent line (-2) and this point, we can write the equation of the tangent line in the point-slope form: $$ y - (-5) = -2(x-2) $$ Simplify the equation: $$ y + 5 = -2(x - 2) $$

Graph the curve and the tangent line

Now that we have the equation of the curve \(y = x^3 - 4x^2 + 2x - 1\) and the equation of the tangent line \(y + 5 = -2(x - 2)\) at the point \((2,-5)\), use graphing software to plot both functions on the same coordinate plane as instructed in the exercise. Remember to adjust the viewing window to see the curve and tangent line clearly.