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y = f 0

In the case of rational function, we can not determine y-intercept if the function is not defined at x=0.

Find y-intercepts of reciprocal function :

y = 1 x

The equation can not be solved for x=0. Thus, reciprocal function does not have y-intercepts.

Find y-intercepts :

f x = 2 x 3

Putting x=0,

f 0 = - 3

Real values of rational function

In order to find real values of rational function for real x, we rearrange the given rational function to form a quadratic equation in x. Let us consider a rational polynomial of degree 2 given by (we mean that the highest degree of polynomials involved in the ratio is 2) :

y = f x = a 1 x 2 + b 1 x + c 1 a 2 x 2 + b 2 x + c 2

Rearranging to form a quadratic equation in x, we have :

y a 2 x 2 + b 2 x + c 2 = a 1 x 2 + b 1 x + c 1

y a 2 a 1 x 2 + y b 2 b 1 x + c 2 y c 1 = 0

For x to be real, D≥0. Hence,

D = y b 2 b 1 2 4 x y a 2 a 1 c 2 y c 1 0

We see that discriminant itself is a quadratic inequality. Depending on the nature of coefficient of “ y 2 ” in the quadratic equation and determinant of the corresponding quadratic equation, the inequality is solved for “y”. This, in turn, allows us to determine the real interval(s) of y corresponding to real x.

We should clearly understand that these are real values of y corresponding to real x. This interval need not be the range of the function. Recall that range of a function contains values of y for values of x in the domain of function – not all real values of x. Now function is not defined for certain values of x, which are zeroes of denominator of the rational function. Hence domain is not the real number set R. This distinction should always be kept in mind.

Problem : Find value of x for which given function has least value.

y = f x = x 2 6 x + 5 x 2 + 2 x + 1

Solution : Rearranging to form a quadratic equation in x, we have :

y x 2 + 2 y x + y = x 2 6 x + 5 y 1 x 2 + 2 y + 3 x + y 5 = 0

For x real, D≥0.

4 y + 3 2 4 y 1 X y 5 0 y 2 + 6 y + 9 y 2 6 y + 5 0 12 y + 4 0 y 1 3

The real values of y, therefore, lies in the interval [-1/3, ∞). The least value of y in the interval of real values is -1/3. We should, however, check that value of y=-1/3 does not correspond to value of x which is not permitted. Here, the denominator polynomial is x 2 + 2 x + 1 = x + 1 2 . Thus, x - 1 . The domain of the function is R-{-1}.

Now, we calculate value of x corresponding to y as :

1 3 = x 2 6 x + 5 x 2 + 2 x + 1

x 2 4 x + 4 = 0 x 2 2 = 0 x = 2

This point belongs to the domain of the function as it is different to excluded value. Hence, least value of y is -1/3.

Problem : For what values of “a”, the function given here assumes all real values for real x.

y = f x = a x 2 + 3 x 4 3 x 4 x 2 + a

Solution : Rearranging to form a quadratic equation in x, we have :

3 y x 4 y x 2 + a y = a x 2 + 3 x 4

a + 4 y x 2 + 3 1 y x a y + 4 = 0

For x real, D≥0.

9 1 y 2 + 4 a + 4 y X a y + 4 0

9 + 16 a y 2 + 4 a 2 + 46 y + 9 + 16 a 0

This is a quadratic inequality in y. This inequality holds when coefficient of “ y 2 ” is positive and discriminant is less than equal to zero i.e. D’≤0.

9 + 16 a > 0

a > - 9 16

and

D = 4 a 2 + 46 2 4 9 + 16 a 2 0 2 a 2 + 23 2 9 + 16 a 2 0 2 a 2 + 23 + 9 + 16 a 2 a 2 + 23 9 16 a 0 2 a 2 + 16 a + 32 2 a 2 16 a + 14 0 a 2 + 8 a + 16 a 2 8 a + 7 0 a + 4 2 a 2 8 a + 7 0 a 2 8 a + 7 0 a 2 8 a + 7 0

The roots of corresponding quadratic equation in a is :

a = 1,7

Hence interval of a satisfying inequality is [1,7]

We have two intervals i.e. a>-9/16 and a [ 1,7 ] corresponding to two simultaneous conditions. Therefore, values of a is intersection of these two intervals :

a [ 1,7 ]

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Source:  OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64
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