Write an equation of the line in slope intercept form that pass through the two

1 answer:

Ber [7]3 years ago

8 0

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -2}}\quad ,&{{ 15}})\quad 
% (c,d)
&({{ 9}}\quad ,&{{ -18}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-18-15}{9-(-2)}\implies \cfrac{-18-15}{9+2}
\\\\\\
\cfrac{-33}{11}\implies -3$

$\bf \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-15=-3[x-(-2)]
\\\\\\
y-15=-3(x+2)\implies y-15=-3x-6\implies y=-3x+9$

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WITCHER [35]

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The domain is the set of allowed x inputs of a function.

The graph shows that x = 0 is not allowed because of the vertical asymptote located here. It seems like any other x value is fine though.

<h3>Domain: set of all real numbers but $x \ne 0$ </h3>

To write this in interval notation, we can say $(-\infty, 0) \cup (0, \infty)$ which is the result of poking a hole at 0 on the real number line.

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The range deals with the y values. The graph makes it seem like it stretches on forever in both up and down directions. If this is the case, then the range is the set of all real numbers.

<h3>Range: Set of all real numbers</h3>

In interval notation, we would say $(-\infty, \infty)$ which is almost identical to the interval notation of the domain, except this time of course we aren't poking at hole at 0.

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Part (b)

<h3>The x intercepts are x = -4 and x = 4</h3>

We can compact that to the notation $x = \pm 4$

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Part (c)

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Part (d)

The vertical asymptote is located at x = 0, so the equation of the vertical asymptote is naturally x = 0. Every point on the vertical dashed line has an x coordinate of zero. The y coordinate can be anything you want.

<h3>Answer: x = 0 is the vertical asymptote</h3>

=======================================================

Part (e)

The oblique or slant asymptote is the diagonal dashed line.

It goes through (0,0) and (2,6)

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If you were to zoom out on the graph (if possible), then you should notice the branches of the hyperbola stretch forever upward but they slowly should approach the "fencing" that is y = 3x. The same goes for the vertical asymptote as well of course.

<h3>Answer: Oblique asymptote is y = 3x</h3>

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