#### Answer

The graph is plotted below. All the details are explained in the Work step by step.

#### Work Step by Step

$$f(x)=x^3+2x\hspace{1cm}x_0=0$$
a) Here, since $x_0=0$, we will plot $y=f(x)$ over the interval $-1/2\le x\le3$.
The graph of $y=f(x)$ is represented by the red curve.
b) and c) With $x_0=0$: $$q(h)=\frac{f(h)-f(0)}{h}=\frac{f(h)-(0^3+2\times0)}{h}=\frac{f(h)}{h}$$
The graph of $q(h)$ is also plotted and represented by the purple line.
As $h\to0$, from the graph, we see that $q$ approaches $2$. Therefore, $$\lim_{h\to0}q(h)=2$$
d) For $x_0=0$: $$y=f(0)+q(h)\times(x-0)=0+q(h)\times x=q(h)\times x$$
- With $h=3$: $$y=q(3)\times x=\frac{f(3)}{3}\times x=\frac{(3^3+2\times3)x}{3}=\frac{33x}{3}=11x$$
This secant line is represented by the black line.
- With $h=2$: $$y=q(2)\times x=\frac{f(2)}{2}\times x=\frac{(2^3+2\times2)x}{2}=\frac{12x}{2}=6x$$
This secant line is represented by the blue line.
- With $h=1$: $$y=q(1)\times x=\frac{f(1)}{1}\times x=\frac{(1^3+2\times1)x}{1}=\frac{3x}{1}=3x$$
This secant line is represented by the green line.
- The tangent line at $x_0=0$:
The tangent line at $x=0$ will have this form: $$y=\lim_{h\to0}q(h)x+m=2x+m$$
To find m, we use the fact that the tangent line is at $x=0$: $$2\times0+m=f(0)=0$$ $$m=0$$
So the tangent line at $x_0$ is $y=2x$. It is represented by the orange line.