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tiny-mole [99]
3 years ago
10

Sketch the graph of each rational function showing all the key features. Verify your graph by graphing the function on

Mathematics
1 answer:
leva [86]3 years ago
5 0

Answer:

Rational Function

Step-by-step explanation:

Gives the function:

f(x) = \frac{x-2}{3x-2x^2}

It has the following,

Domain

x∈ \mathbb{R} : x\neq 0, x\neq 3/2

Range

f \in \mathbb{R} : f \leq -1, f \geq -\frac{1}{9}

End behaviour

As xx \rightarrow \infty, y \rightarrow 0

Horizontal and Vertical Asymptote

Vertical x = 0, x=3/2

Horizontal y = 0

<em>x-intercept: (2,0)</em>

<em>y-intercept: none</em>

<em>**The Graph is in the annex</em>

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Find the length of JL<br> 15<br> None of the other answers are correct<br> 12<br> 3
r-ruslan [8.4K]

Answer:

D: 12

Step-by-step explanation:

The entire segment KI is composed of the segments KJ and JI. So:

KI=KJ+JI

We know that KI is 15.

KJ is (2x-15) and JI is (2x-6). Substitute:

15=(2x-15)+(2x-6)

Solve for x. Combine like terms:

15=4x-21

Add 21 to both sides:

4x=36

Hence, the value of x is:

x=9

We know that JI is 2x-6.

We will substitute 9 for x to find JI. So:

JI=2(9)-6=18-6=12

Hence, our answer is D.

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3 years ago
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Answer:

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Step-by-step explanation:

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Expand each expressions and combine like terms if possible<br>3/4(16x – 27 – 1)​
VLD [36.1K]

Answer:

12x-21

Step-by-step explanation:

3/4(16x-27-1)

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Use the denition of the derivative to find f 0 (3), where f (x) = 3x+5 / 2x−1
Crazy boy [7]

Answer:

f'(3)= -\frac{13}{25}

Step-by-step explanation:

We are asked to find f'(3) of function f(x)=\frac{3x+5}{2x-1} using definition of derivatives.

Limit definition of derivatives:

f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

Let us find f(3+h) and f(3).

f(3+h)=\frac{3(3+h)+5}{2(3+h)-1}

f(3+h)=\frac{9+3h+5}{6+2h-1}\\\\f(3+h)=\frac{3h+14}{2h+5}

f(3)=\frac{3(3)+5}{2(3)-1}

f(3)=\frac{9+5}{6-1}\\\\f(3)=\frac{14}{5}

Substituting these values in limit definition of derivatives, we will get:

f'(3)= \lim_{h \to 0} \frac{f(3+h)-f(3)}{h}

f'(3)= \lim_{h \to 0} \frac{\frac{3h+14}{2h+5}-\frac{14}{5}}{h}

Make a common denominator:

f'(3)= \lim_{h \to 0} \frac{\frac{(3h+14)*5}{(2h+5)*5}-\frac{14(2h+5)}{5(2h+5)}}{h}

f'(3)= \lim_{h \to 0} \frac{\frac{5(3h+14)-14(2h+5)}{5(2h+5)}}{h}

f'(3)= \lim_{h \to 0} \frac{5(3h+14)-14(2h+5)}{5h(2h+5)}

f'(3)= \lim_{h \to 0} \frac{15h+70-28h-70}{5h(2h+5)}

f'(3)= \lim_{h \to 0} \frac{-13h}{5h(2h+5)}

Cancel out h:

f'(3)= \lim_{h \to 0} \frac{-13}{5(2h+5)}

f'(3)= \frac{-13}{5(2(0)+5)}

f'(3)= \frac{-13}{5(5)}

f'(3)= -\frac{13}{25}

Therefore, f'(3)= -\frac{13}{25}.

8 0
3 years ago
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