Chapter 12.3, Problem 78E

(a)

To determine

To sketch: the graph of the function $f\left(x\right)=3x^2-2x$ .

Explanation of Solution

Graph:

The graph of the function $f\left(x\right)=3x^2-2x$ are shown in Figure 1.

  

Interpretation:

From Figure 1, it is observed that the shape of the function f is parabola open upward.

(b)

To determine

To find: the coordinates of the vertex of the parabola $f\left(x\right)=3x^2-2x$ by using the trace feature to approximate.

Answer to Problem 78E

The coordinates of the vertex of the parabola are $x=\frac{1}{3}\text{ and }y=-\frac{1}{3}$ .

Explanation of Solution

Interpretation:

From Figure 1, it is observed that the parabola is upward open.

Note that, the minimum value of the upward open parabola is vertex point.

The function $f\left(x\right)=3x^2-2x$ is increasing on the interval $\left(-\infty ,\frac{1}{3}\right)$ and the function is increasing on the interval $\left(\frac{1}{3},\infty \right)$ and the turning point of the function $f\left(x\right)=3x^2-2x$ at $x=\frac{1}{3}$ .

The minimum value of $f\left(x\right)=3x^2-2x$ is $-\frac{1}{3}$ at $x=\frac{1}{3}$ .

Therefore, the vertex of the parabola is $\left(\frac{1}{3},-\frac{1}{3}\right)$ .

Hence, the coordinates of the vertex of the parabola are $x=\frac{1}{3}\text{ and }y=-\frac{1}{3}$ .

(c)

To determine

To find: the slope of the tangent of function $f\left(x\right)=3x^2-2x$ at vertex.

Answer to Problem 78E

The slope of the tangent of function f at vertex is 0.

Explanation of Solution

Calculation:

From part (a), the derivative of the function $f\left(x\right)=3x^2-2x$ is $f^{\prime }\left(x\right)=6x-2$ ,

Note that, the slope of the tangent of the graph of the function $f\left(x\right)=3x^2-2x$ at $\left(x_0,f\left(x_0\right)\right)$ is the derivative of the function $f\left(x\right)=3x^2-2x$ at $x=x_0$ .

The slope of the tangent of function $f\left(x\right)=3x^2-2x$ at $\left(\frac{1}{3},-\frac{1}{3}\right)$ is computed as follows,

Substitute $x=\frac{1}{3}$ in $f^{\prime }\left(x\right)=6x-2$ ,

  $\begin{array}{c}{f}^{\prime }\left(\frac{1}{3}\right)=6\left(\frac{1}{3}\right)-2\\ =2-2\\ =0\end{array}$

Therefore, the slope of the tangent of function f at vertex is 0.

(d)

To determine

To make: the conjecture about the slope of the tangent line at the vertex of an arbitrary parabola.

Answer to Problem 78E

The slope of the tangent line at vertex of the any parabola is zero.

Explanation of Solution

Interpretation:

From part (c), the slope of the tangent line of the parabola at vertex is zero.

The vertex point is a minimum point of upward open parabola and maximum point of downward open parabola. Also, the vertex point is a turning point only it never gets sharp nose.

The tangent of the vertex point is horizontal line and the slope of the horizontal line is zero.

Therefore, the slope of the tangent line at vertex of the any parabola is zero.