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2 years ago

In Lesson 3.05 we discussed comparing the key features of two functions given in different forms. Given the functions below:

Mathematics

2 answers:

Tamiku [17]2 years ago

8 0

Answer:

A. f(x): $(0,\frac{-1}{3})$ , g(x): (0,0)

B. f(x): 'y= 0' as horizontal asymptote, g(x): 'x= 4' as vertical asymptote.

Step-by-step explanation:

We are given the functions, $f(x)=\frac{1}{x-3}$ and graph of g(x).

Part A: We know that,

'Y-intercepts are the points where the graph of the function crosses y-axis'.

That is, 'y-intercept is obtained when x= 0'.

So, we have,

When x= 0, $f(0)=\frac{1}{0-3}$ i.e. $f(0)=\frac{-1}{3}$ .

Thus, the y-intercept of function f(x) is the point $(0,\frac{-1}{3})$ .

Also, from the graph of g(x), we see that, the graph crosses y-axis at the point (0,0).

Thus, the y-intercept of g(x) is (0,0).

Part B: We know that,

'Asymptotes are the line or curves approaching the graphs of the functions arbitrarily close'.

Now, as we have the function f(x) is a rational function and the degree of numerator is less than the degree of the denominator.

So, the function f(x) will have 'y= 0' as the horizontal asymptote.

Also, from the graph of the function g(x), we see that,

The line 'x= 4' is the vertical asymptote as it is approaching the graph closely.

So, the function g(x) have 'x= 4' as the vertical asymptote.

Stels [109]2 years ago

5 0

Try desmos or ask the teacher

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The vector (a) is a multiple of the vector (2i +3j) and (b) is a multiple of (2i+5j) The sum (a+b) is a multiple of the vector (

kow [346]

Answer:

$\|a\| = 5\sqrt{13}$ .

$\|b\| = 3\sqrt{29}$ .

Step-by-step explanation:

Let $m$ , $n$ , and $k$ be scalars such that:

$\displaystyle a = m\, (2\, \vec{i} + 3\, \vec{j}) = m\, \begin{bmatrix}2 \\ 3\end{bmatrix}$ .

$\displaystyle b = n\, (2\, \vec{i} + 5\, \vec{j}) = n\, \begin{bmatrix}2 \\ 5\end{bmatrix}$ .

$\displaystyle (a + b) = k\, (8\, \vec{i} + 15\, \vec{j}) = k\, \begin{bmatrix}8 \\ 15\end{bmatrix}$ .

The question states that $\| a + b \| = 34$ . In other words:

$k\, \sqrt{8^{2} + 15^{2}} = 34$ .

$k^{2} \, (8^{2} + 15^{2}) = 34^{2}$ .

$289\, k^{2} = 34^{2}$ .

Make use of the fact that $289 = 17^{2}$ whereas $34 = 2 \times 17$ .

$\begin{aligned}17^{2}\, k^{2} &= 34^{2}\\ &= (2 \times 17)^{2} \\ &= 2^{2} \times 17^{2} \end{aligned}$ .

$k^{2} = 2^{2}$ .

The question also states that the scalar multiple here is positive. Hence, $k = 2$ .

Therefore:

$\begin{aligned} (a + b) &= k\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 2\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 16\, \vec{i} + 30\, \vec{j}\\ &= \begin{bmatrix}16 \\ 30 \end{bmatrix}\end{aligned}$ .

$(a + b)$ could also be expressed in terms of $m$ and $n$ :

$\begin{aligned} a + b &= m\, (2\, \vec{i} + 3\, \vec{j}) + n\, (2\, \vec{i} + 5\, \vec{j}) \\ &= (2\, m + 2\, n) \, \vec{i} + (3\, m + 5\, n) \, \vec{j} \end{aligned}$ .

$\begin{aligned} a + b &= m\, \begin{bmatrix}2\\ 3 \end{bmatrix} + n\, \begin{bmatrix} 2\\ 5 \end{bmatrix} \\ &= \begin{bmatrix}2\, m + 2\, n \\ 3\, m + 5\, n\end{bmatrix}\end{aligned}$ .

Equate the two expressions and solve for $m$ and $n$ :

$\begin{cases}2\, m + 2\, n = 16 \\ 3\, m + 5\, n = 30\end{cases}$ .

$\begin{cases}m = 5 \\ n = 3\end{cases}$ .

Hence:

$\begin{aligned} \| a \| &= \| m\, (2\, \vec{i} + 3\, \vec{j})\| \\ &= m\, \| (2\, \vec{i} + 3\, \vec{j}) \| \\ &= 5\, \sqrt{2^{2} + 3^{2}} = 5 \sqrt{13}\end{aligned}$ .

$\begin{aligned} \| b \| &= \| n\, (2\, \vec{i} + 5\, \vec{j})\| \\ &= n\, \| (2\, \vec{i} + 5\, \vec{j}) \| \\ &= 3\, \sqrt{2^{2} + 5^{2}} = 3 \sqrt{29}\end{aligned}$ .

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Answer:

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