Question

In Exercises \(9-12,\) at the indicated point find (a) the slope of the curve, (b) an equation of the tangent, and (c) an equation of the tangent. (d) Then draw a graph of the curve, tangent line, and normal line in the same square viewing window. $$y=x^{2}-4 x\( at \)x=1$$

Short answer

The slope of the curve at the point x=1 is -2. The equation of the tangent line at x=1 is \(y = -2x + 5\) and the equation of the normal line at this point is \(y = \frac{1}{2}x - \frac{7}{2}.\)

Step-by-step solution

Calculating the Derivative of the Function

To find the slope of the curve at a given point, we first need to calculate the derivative of the function. Start by differentiating the function \(y = x^{2} - 4x\). Using the power rule \(\frac{d}{dx}(x^n) = nx^{n-1}\) and the sum/difference rule, the derivative of the function \(y'\) is given by \(y' = 2x - 4.\)

Determine the Slope of the Curve

Substitute \(x = 1\) into \(y' = 2x - 4\) to get the slope of the curve at the point \(x = 1\). Thus, the slope at \(x = 1\) is \(2*1 - 4 = -2.\)

Find an Equation of the Tangent

We now need to determine the value of \(y\) at \(x = 1\) to get the coordinates of the point of tangency. Substituting \(x = 1\) into \(y = x^{2} - 4x\) gives \(y = 1^{2} - 4*1 = -3.\) So, the point of tangency is \((1, -3).\) An equation of the tangent line can be obtained using the point-slope form of the linear equation, which states that for any point \((x1, y1),\) the equation of the line is \(y - y1 = m(x - x1),\) where \(m\) is the slope of the line. The equation of the tangent line at \(x = 1\) is thus \(y + 3 = -2(x - 1)\) which simplifies to \(y = -2x + 5.\)

Find an Equation of the Normal

The normal line to a curve at a given point is the line perpendicular to the tangent at that point. In a plane, two lines are perpendicular if the product of their slopes is -1. The slope of the normal line is hence given by the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is -2, the slope of the normal line is \(\frac{1}{2}.\) Substituting into the point-slope form, the equation of the normal line at the point (1, -3) is \(y + 3 = \frac{1}{2}(x - 1)\) which simplifies to \(y = \frac{1}{2}x - \frac{7}{2}.\)

Drawing the Graph of the Function, Tangent, and Normal line

To depict the graph, plot the quadratic function \(y = x^{2} - 4x,\) the tangent line \(y = -2x + 5,\) and the normal line \(y = \frac{1}{2}x - \frac{7}{2}\) in the same square viewing window. The given point (1, -3) should appear as the point of tangency on the curve. The tangent line touches the curve at this point and is distinctly slope downwards while the normal line intersects the function and the tangent line at this point with a noticeably slower incline.