Question

homework4. problem 1 (1 point) let (f = x^{2}-3x - 1) and find the values below 1. (f(x + h)=) 2. ((f(x + h)-f(x))=) 3. (lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=) 4. find the equation of the line tangent to the graph of (f) at (x=-2). (y=)

Explanation:

Step1: Substitute \(x + h\) into \(f(x)\)

\[

$$\begin{align*} f(x + h)&=(x + h)^2-3(x + h)-1\\ &=x^{2}+2xh+h^{2}-3x - 3h-1 \end{align*}$$

\]

Step2: Calculate \(f(x + h)-f(x)\)

\[

$$\begin{align*} f(x + h)-f(x)&=(x^{2}+2xh+h^{2}-3x - 3h-1)-(x^{2}-3x - 1)\\ &=x^{2}+2xh+h^{2}-3x - 3h-1 - x^{2}+3x + 1\\ &=2xh+h^{2}-3h\\ &=h(2x + h-3) \end{align*}$$

\]

Step3: Find the limit

\[

$$\begin{align*} \lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}&=\lim_{h ightarrow0}\frac{h(2x + h-3)}{h}\\ &=\lim_{h ightarrow0}(2x + h-3)\\ &=2x-3 \end{align*}$$

\]

Step4: Find the slope and a point for the tangent - line

When \(x=-2\), the slope \(m = 2(-2)-3=-4 - 3=-7\).
When \(x = - 2\), \(f(-2)=(-2)^{2}-3(-2)-1=4 + 6-1 = 9\).
Using the point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(-2,9)\) and \(m=-7\), we have \(y - 9=-7(x + 2)\).
\[

$$\begin{align*} y-9&=-7x-14\\ y&=-7x-5 \end{align*}$$

\]

Answer:

1. \(x^{2}+2xh+h^{2}-3x - 3h-1\)
2. \(h(2x + h-3)\)
3. \(2x-3\)
4. \(y=-7x - 5\)