homework4. problem 1 (1 point) let (f = x^{2}-3x - 1) and find the values below 1. (f(x + h)=) 2. ((f(x +…

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\rightarrow0}\frac{f(x + h)-f(x)}{h}=) 4. find the equation of the line tangent to the graph of (f) at (x=-2). (y=)

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\rightarrow0}\frac{f(x + h)-f(x)}{h}=) 4. find the equation of the line tangent to the graph of (f) at (x=-2). (y=)

Answer

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\rightarrow0}\frac{f(x + h)-f(x)}{h}&=\lim_{h\rightarrow0}\frac{h(2x + h-3)}{h}\ &=\lim_{h\rightarrow0}(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)