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Sophie [7]
3 years ago
13

If the function y=1/x is translated to the left 2 units and up 6 units, what is the new horizontal asymptote

Mathematics
1 answer:
Aliun [14]3 years ago
5 0

Answer:

y=6

Step-by-step explanation:

The horizontal asymptote for y = 1/x is the x-axis, y = 0.

Shifting the graph up 6 makes the asymptote move up, too, to y = 6.

See the attached image:

y = 1/x is purple

y = 1/(x + 2) + 6 is black

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The maximum value of the function: f(x)= -5 x ^2 +30x-200 is?
VARVARA [1.3K]

Answer:

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Step-by-step explanation:

we are given a quadratic function

\displaystyle f(x) =  - 5 {x}^{2}  + 30x - 200

we want to figure out the minimum value of the function

to do so we need to figure out the minimum value of x in the case we can consider the following formula:

\displaystyle x _{ \rm  min} =  \frac{ - b}{2a}

the given function is in the standard form i.e

\displaystyle f(x) = a {x}^{2}  + bx + c

so we acquire:

  • b=30
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thus substitute:

\displaystyle x _{ \rm  min} =  \frac{ - 30}{2. - 5}

simplify multiplication:

\displaystyle x _{ \rm  min} =  \frac{ - 30}{ - 10}

simply division:

\displaystyle x _{ \rm  min} =  3

plug in the value of minimum x to the given function:

\displaystyle f (3)=  - 5 {(3)}^{2}  + 30.3 - 200

simplify square:

\displaystyle f (3)=  - 5 {(9)}^{}  + 30.3 - 200

simplify multiplication:

\displaystyle f (3)=  - 45  + 90- 200

simplify:

\displaystyle f (3)=   - 155

hence,

the minimum value of the function is -155

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Now we can simplify 42/99 to 14/33

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