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3 years ago

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Which function does not have a vertical asymptote

1 answer:

Gemiola [76]3 years ago

8 0

Answer:

$y=\frac{5x}{1+2x^2}$

Step-by-step explanation:

$Option ~A ~ does ~ not ~ have ~a ~ vertical~ asymptote$

<u>--------------------------</u>

Hope it helps...

Have a great day!!

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Match each transformation of function fwith a feature of the transformed function.

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The function f(x) has its vertical asymptote at x = 0
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<h3>How to match the functions and their properties?</h3>

The equations of the functions are given as:

f(x) = ln(x)

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2 years ago

Help please<3 for all questions with solutions plz

antiseptic1488 [7]

Answer:

a) $h = -\frac{7}{4}$ , b) $h = f\,\circ\,g (-5) = \frac{3}{2}$ , c) $h = g\,\circ\,f (-2) = \frac{3}{8}$

Step-by-step explanation:

We proceed to solve each exercise below:

a) $f(-2) +3\cdot g(0)$

$h = \frac{2\cdot (-2)}{3\cdot (-2)+5}+3\cdot \left(\frac{3}{0+4} \right)$

$h = \frac{-4}{-6+5} +3\cdot \left(\frac{3}{4} \right)$

$h = -4+\frac{9}{4}$

$h = \frac{-16+9}{4}$

$h = -\frac{7}{4}$

b) $h = f\,\circ \,g (-5)$

$h = f\,\circ\,g (x) = \frac{\frac{6}{x+4} }{\frac{9}{x+4}+5 }$

$h = f\,\circ\,g (x)= \frac{\frac{6}{x+4} }{\frac{9+5\cdot (x+4)}{x+4} }$

$h = f\,\circ\,g (x) = \frac{6}{29+5\cdot x}$

$h = f\,\circ\,g (-5) = \frac{6}{29+5\cdot (-5)}$

$h = f\,\circ\,g (-5) = \frac{3}{2}$

c) $h = g\,\circ\,f(-2)$

$h = g\,\circ \, f (x) = \frac{3}{\frac{2\cdot x}{3\cdot x + 5} + 4 }$

$h = g\,\circ\,f (x) = \frac{3}{\frac{2\cdot x +4\cdot (3\cdot x +5)}{3\cdot x +5 } }$

$h = g\,\circ \,f (x) = \frac{3\cdot (3\cdot x +5)}{2\cdot x +12\cdot x +20}$

$h = g\,\circ\,f(x) = \frac{9\cdot x +15}{14\cdot x +20}$

$h = g\,\circ\,f(-2) = \frac{9\cdot (-2)+15}{14\cdot (-2)+20}$

$h = g\,\circ\,f (-2) = \frac{-3}{-8}$

$h = g\,\circ\,f (-2) = \frac{3}{8}$

6 0

3 years ago