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2 years ago

9

An example of a non-linear and linear function?

2 answers:

ollegr [7]2 years ago

8 0

Answer:

Non-linear: y=x^(2)+3x-17

Linear: y= 8x+1

Step-by-step explanation:

Quadratic, exponential, and square root functions are all non-linear. They should not be straight. There is always an x and/or y value that they will never be on. Linear functions can touch all values. That sounds very confusing, but it's better than nothing, so I'm sorry.

I don't know a better way to explain it.

lesya [120]2 years ago

5 0

Linear functions have a constant slope, so nonlinear functions have a slope that varies between points. Algebraically, linear functions are polynomials with highest exponent equal to 1 or of the form y = c where c is constant. Nonlinear functions are all other functions. An example of a nonlinear function is y = x^2.

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3 years ago

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2 years ago

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3 years ago

F(x) = x^2+x-2/x^2-3x-4

Andrej [43]

i. Domain and Range

The given function is

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

The domain of this function is,

$x^2-3x-4\ne 0$

$(x-4)(x+1)\ne 0$

$x\ne4,xne -1$

The range refers to the y-values for which x is defined. x is defined for all values of y.

The range is all real numbers. See graph

ii. x-and-y-intercept

For x- intercept intercept we put $f(x)=0$

This implies that;

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

This will give us

$x^2+x-2=0$

$\Rightarrow x^2+x-2=0$

$\Rightarrow x^2+2x--x-2=0$

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow (x+2)(x-1)=0$

$\Rightarrow (x+2)=0,(x-1)=0$

$\Rightarrow x=-2,x=1$

The x-intercepts are $(-2,0),(1,0)$

For y-intercept, we put

$x=0$ to obtain;

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

$f(0)=\frac{1}{2}$

The y-intercept is

$(0,\frac{1}{2})$

iii. Horizontal asyptote

Since degree of the numerator and the denominator are the same, there is a horizontal asymptote

To find the horizontal asymptote.

We divide the leading coefficient of the numerator by the leading coefficient of the denominator.

The horizontal asymptote is $y=\frac{1}{1}=1$

iv. Vertical asymptote

To find the vertical asymptote, we equate the denominator to zero to get;

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

This implies that;

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

Split the middle term

$x(x+1)-4(x+1)=0$

Factor

$(x+1)(x-4)=0$

Factor further

$(x+1)=0,(x-4)=0$

$x=-1,x=4$

The vertical asymptotes are $x=-1,x=4$

8 0

2 years ago

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