Problem 29 - Calculus Early Transcendentals

Question

Tangent lines with zero slope a. Graph the function \(f(x)=x^{2}-4 x+3\) b. Identify the point \((a, f(a))\) at which the function has a tangent line with zero slope. c. Confirm your answer to part (b) by making a table of slopes of secant lines to approximate the slope of the tangent line at this point.

Step-by-Step Solution

Answer

Answer: The tangent line has a zero slope at the point (2, -1).

Step 1: Graph the function

To graph the function \(f(x) = x^2 - 4x + 3\), plot it on a coordinate plane to visualize the shape of the graph. You can either do this by hand or use graphing software. The graph will have the shape of a parabola that opens upwards.

Step 2: Find the point with zero slope

To find the point \((a, f(a))\) where the tangent has a zero slope, we need to find the critical point of the function. In this case, the first derivative of the function represents the slope of the tangent line at any given point in the graph: First, find the first derivative of \(f(x)\): \[ f'(x) = \frac{d}{dx}(x^2 - 4x + 3) = 2x - 4 \] Now, set the first derivative equal to zero and solve for \(x\): \[ 2x - 4 = 0 \Rightarrow x = 2 \] So, the \(x\)-coordinate of the point is \(2\). Now, let's find the \(y\)-coordinate: \[ f(2) = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1 \] So, the point where the tangent line has a zero slope is \((2, -1)\).

Step 3: Confirm the answer with a table of slopes

We can confirm the answer by creating a table of slopes of secant lines near the point \((2, -1)\). We can calculate the slope of the secant line between \((2, -1)\) and \((x, f(x))\): The slope of a secant line is given by: \[ m(x) = \frac{f(x)-f(2)}{x-2} \] The table should look like this: | \(x\) | \(f(x)\) | \(m(x)\) | |-----|-------|-------| | 1.9 | -0.99 | -0.100 | | 1.99| -1.0001| -0.010 | | 2 | -1 | - | | 2.01| -1.0001| 0.010 | | 2.1 | -0.99 | 0.100 | In the table, as \(x\) approaches 2, the slope (\(m(x)\)) is getting close to zero. This confirms that at the point \((2, -1)\), the function has a tangent line with a zero slope.