How to find the equation of vertical and horizontal asymptotes?

1 answer:

amm18123 years ago

5 0

Please find some specific examples of functions for which you want to find vert. or horiz. asy. and their equations. This is a broad topic.

Very generally, vert. asy. connect only to rational functions; if the function becomes undef. at any particular x-value, that x-value, written as x = c, is the equation of one vertical asy.

Very generally, horiz. asy. pertain to the behavior of functions as x grows increasingly large (and so are often associated with rational functions). To find them, we take limits of the functions, letting x grow large hypothetically, and see what happens to the function. Very often you end up with the equation of a horiz. line, your horiz. asy., which the graph usually (but not always) does not cross.

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East of Star, the railroad tracks intersect highway 23o as shown above. If 1=133 , find 2.

GrogVix [38]

Hi there.

These two angles sum up to be 180 degrees, so they are supplementary angles. Now, since these angles sum up to 180 degrees (to form a straight line), we only have to subtract 133 from 180 to find angle 2's measurement.

180 - 133 = 47

Angle 2's measurement is 47 degrees.

Voila!

I hope this helps! Have a great day. :)

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3 years ago

-1/225x^2+2/3x what is the vertex of this quadratic function?

Julli [10]

$\bf \textit{vertex of a parabola}\\ \quad \\\\

\begin{array}{lccclll}
y=&-\frac{1}{225}x^2&+\frac{2}{3}x\\\\
y=&-\frac{1}{225}x^2&+\frac{2}{3}x&+0\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
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3 years ago

Plz help<br>urgent !!!!<br>will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

$\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}$ = $\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}$ = $\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}$ = $\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}$

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...