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3 years ago

What are the vertical and horizontal asymptotes for the function f(x)= 3x2/x2-4

Mathematics

1 answer:

Alecsey [184]3 years ago

3 0

Answer: f(x) will have vertical asymptotes at x=-2 and x=2 and horizontal asymptote at y=3.

Step-by-step explanation:

Given function: $f(x)=\dfrac{3x^2}{x^2-4}$

The vertical asymptote occurs for those values of x which make function indeterminate or denominator 0.

i.e. $x^2-4=0\Rightarrow\ x^2=4\Rightarrow\ x=\pm2$

Hence, f(x) will have vertical asymptotes at x=-2 and x=2.

To find the horizontal asymptote , we can see that the degree of numerator and denominator is same i.e. 2.

So, the graph will horizontal asymptote at $y=\dfrac{\text{Coefficient of }x^2\text{ in numerator}}{\text{Coefficient of }x^2\text{ in denominator}}$

i.e. $y=\dfrac{3}{1}=3$

Hence, f(x) will have horizontal asymptote at y=3.

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The equation (-2) (6x - 5)

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Answer:

Step-by-step explanation:

Givens

(x - 1)(6x - 5) = 0

Solution

Either one of these can be made into 0 by solving the binomials in the brackets.

x - 2 = 0 Add 2 to both sides

x -2 +2= 0+2 Combine

x = 2

6x - 5 = 0 Add 5

6x -5+5 = 0+5 Combine

6x = 5 Divide by 6

6x/6=5/6

x = 5/6

Answer

Integer Answer: x = 2

Fractional Answer x = 5/6

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1 year ago

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Answer:

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y = x^2 +10x -9

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3 years ago

How to find the area of congruent shapes

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you multiply length times width times height.

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3 years ago

Read 2 more answers

Which statement describes the inverse of m(x) = x2 – 17x?

stealth61 [152]

Answer:

The correct option is;

$The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}$

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

$x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

$x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}$

Which can be simplified as follows;

$x = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2} \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}$

And further simplified as follows;

$x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}$

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

$m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}$

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

$The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}$ to increase m⁻¹(x) above the minimum.

8 0

2 years ago

Find the volume of the composite figure. Give your answer in terms of pi.

Sloan [31]

Volume of cone= 1/3 πr²*h
15²=10²+h²
h²=15²-10²=125
h=√125
Volume of cone= 1/3 *π*100*√125=π/3*(100*5√5)=500π√5/3
Volume of sphere=4/3*π*r³=4/3*π*1000=4000π/3
Volume of half sphere=(4000π/3)/2=2000π/3
All Volume=500π√5/3+2000π/3

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3 years ago