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}-3 x-1 \quad\( at \)\quad x=0$$

Short answer

At x=0, the slope of the curve is -3. The equation of the tangent line is \(y=-3x-1\) and the equation of the normal line is \(y=1/3x-1\).

Step-by-step solution

Find the derivative of the function

We're given the function \(y=x^{2}-3 x-1\). To find the slope of the curve at a given point, we need to find the derivative of this function: \(y'=2x - 3\)

Substitute the value of x into the derivative

Substitute the value \(x=0\) into the derivative to find the slope of the tangent at this point. Thus, \(y'(0) = 2(0) - 3 = -3\). This is the slope of the tangent line at the point where \(x=0\).

Use the point-slope form to find the equation of the tangent

The point-slope form of a line is given by \(y-y_{1}=m(x-x_{1})\), where \(m\) is the slope and \((x_{1}, y_{1})\) is a point on the line. We have a point on the tangent line, which is when \(x=0\), the \(y\) value is \(y = (0)^{2}-3(0)-1 = -1\). So, our point is \(0,-1\). The slope of the line (m), as we calculated in the previous step, is \(-3\). Substituting all this into the point slope form, we find the equation of the tangent line to be \(y+1=-3(x-0)\) – which simplifies to \(y=-3x-1\)

Find the equation of the normal line to the curve at the same point

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is \(1/3\). Using this slope and the same point from step 3 in the point slope form, we find the equation of the normal line to be \(y+1=1/3(x-0)\) - which simplifies to \(y=1/3x-1\).

Plot the function, tangent line, and normal line

For this final step, use a graphing tool to plot the function \(y=x^{2}-3 x-1\), the tangent line \(y=-3x-1\), and the normal line \(y=1/3x-1\) on the same graph.