## Alberta Free Tutoring And Homework Help For Math 30-1

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### consider the function(2x^2+x-1)/(x^2-2x-3)FindAsymptotesPoint of DiscontinuityDomain Range

4 years ago The funtion can be factored both in the numerator and denominator. First, doing this in the numerator:

$2x^2+x-1=\left(2x-1\right)\left(x+1\right)$2x2+x1=(2x1)(x+1)

Factoring the denominator, we get:

$x^2-2x-3=\left(x-3\right)\left(x+1\right)$x22x3=(x3)(x+1)

Hence, the function, in its factored form is:

$f\left(x\right)=\frac{\left(2x-1\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}$ƒ (x)=(2x1)(x+1)(x3)(x+1)

In order to find out the asymptotes, we have to identify horizontal or vertical asymptotes. To find out horizontal asymptotes, we can follow this rule that applies to all radical functions:

- If both polynomials are the same degree in the numerator and denominator, divide the coefficients of the highest degree terms.

- If the polynomial in the numerator is a lower degree than the denominator, the x-axis (y = 0) is the horizontal asymptote.

- If the polynomial in the numerator is a higher degree than the denominator, there is no horizontal asymptote.

In our case, the degrees of the numerator and denominator are both 2. Hence, the only horizontal asymptote will be obtained dividing the term of highest degree in the numerator (2) by the corresponding in the denominator (1):

$y=\frac{2}{1}=2$y=21 =2

To identify the vertical asymptotes, we need to check the non-permissible values. They are, in this case:

$x=3;x=-1$x=3;x=1

These are the values for x that we cannot calculate the function, because we would be dividing by zero. If we plot this funtion in a graphing calculator, we will se that only x=3 is the vertical asymptote, because the funtion can be simplified to (cancelling the terms (x-1) on top and on the bottom):

$f\left(x\right)=\frac{2x-1}{x-3}$ƒ (x)=2x1x3

At x=-1, we cannot calculate the function but it doesn't exibit a vertical asymptote behaviour. Hence, this is the point of discontinuity. If we use the formula above to calculate the y-value of this point, we get:

$f\left(-1\right)=\frac{\left[2\left(-1\right)-1\right]}{-1-3}=\frac{3}{4}$ƒ (1)=[2(1)1]13 =34

The domain of the funtion will be all the real numbers except the ones that make a division by zero. Hence:

Domain: {x element IR; x $\ne$ -1,3}

The range of the function would be "y" is an element of the real numbers if it weren't for the horizontal asymptote (y=2). Hence, the range will be:

Range: {y element IR; y $\ne$ 2}

Just to summarize:

Asymptotes: x=3 ; y=2

Point of discontinuity: x=-1, y=3/4

Domain: {x element R; x $\ne$ -1,3}

Range: {y element R; y $\ne$ 2}