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3 years ago

Can anybody help me please

Mathematics

2 answers:

FrozenT [24]3 years ago

6 0

4x + 5y = 13

Is (2, 1) a solution to the given linear equation?

4(2) + 5(1)

8 + 5 = 13

(2, 1) is a solution to the given linear equation

Is (-3, 0) a solution to the given linear equation

4(-3) + 5(0)

-12 + 0 = -12

(-3,0) is not a solution to the given linear equation

Is (0, 1) a solution to the given linear equation

4(0) + 5(1)

0 + 5 = 5

(0, 1) is not a solution to the given linear equation

Hope this helps :)

Alexus [3.1K]3 years ago

3 0

Answer:

The answers follow in order, yes, no, and no.

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=f%28x%29%20-%20%5Cfrac%7Bx%5E%7B2%7D-4%20%7D%7Bx%5E%7B4%7D%20%2Bx%5E%7B3%7D%20-4x%5E%7B2%7D-4%

Llana [10]

a) The given function is

$f(x)=\frac{x^2-4}{x^4+x^3-4x^2-4}$

The domain refers to all values of x for which the function is defined.

The function is defined for

$x^4+x^3-4x^2-4\ne0$

This implies that;

$x\ne -2.69,x\ne 1.83$

b) The vertical asymptotes are x-values that makes the function undefined.

To find the vertical asymptote, equate the denominator to zero and solve for x.

$x^4+x^3-4x^2-4=0$

This implies that;

$x= -2.69,x=1.83$

c) The roots are the x-intercepts of the graph.

To find the roots, we equate the function to zero and solve for x.

$\frac{x^2-4}{x^4+x^3-4x^2-4}=0$

$\Rightarrow x^2-4=0$

$x^2=4$

$x=\pm \sqrt{4}$

$x=\pm2$

The roots are $x=-2,x=2$

d) The y-intercept is where the graph touches the y-axis.

To find the y-inter, we substitute;

$x=0$ into the function

$f(0)=\frac{0^2-4}{0^4+0^3-4(0)^2-4}$

$f(0)=\frac{-4}{-4}=1$

e) to find the horizontal asypmtote, we take limit to infinity

$lim_{x\to \infty}\frac{x^2-4}{x^4+x^3-4x^2-4}=0$

The horizontal asymtote is $y=0$

f) The greatest common divisor of both the numerator and the denominator is 1.

There is no common factor of the numerator and the denominator which is at least a linear factor.

Therefore the function has no holes.

g) The given function is a proper rational function.

There is no oblique asymptote.

See attachment for graph.

6 0

3 years ago