# Intercept of rational function

An intercept of rational function is a point where the graph of the rational function cuts the X- or Y-axis. The rational function is in the form of fraction.$y=\frac{f(x)}{g(x)}, g(x)\neq 0 $

In the rational function , if the numerator never be zero then the graph never touches the x-axis.

The point where graph intersect the X-axis for that y-coordinate is zero and the point where the graph intersect the Y-axis for that x-coordinate is zero.

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Steps to find the intercept of rational function

**To find x-intercept :**

1) Plug in y=0

2) Find the value of x. Sometimes directly,sometimes by factoring.

**To find y-intercept :**

1) Plug in x = 0

2) Find the value of y.

**Examples:**

1) Find the x-intercept of the $y=\frac{(x-2)(x+1)}{(x-3)}$ rational function.

**Solution:**$y=\frac{(x-2)(x+1)}{(x-3)}$

To find x-intercept , plug in y= 0

0 =$y=\frac{(x-2)(x+1)}{(x-3)}$

So, (x-2)(x+1) = 0 ( cross multiply)

x-2 =0 and x+1 = 0

x= 2 and x = -1

∴ x-intercepts are (2,0) and (-1,0)

2) Find the x-intercept of $y = y=\frac{(x^{2}-5x + 6)}{(x-1)}$.

**Solution :**$y=\frac{(x^{2}-5x + 6)}{(x-1)}$

To find x-intercept , plug in y= 0

0 =$y=\frac{(x^{2}-5x + 6)}{(x-1)}$

So, $y=x^{2}-5x + 6$ = 0 ( cross multiply)

Now the equation is quadratic, so we will find the factors of it.

(x-3)(x-2)=0

x-3 =0 and x-2 = 0

x= 3 and x = 2

∴ x-intercepts are (3,0) and (2,0)

3) Find the y-intercept of $y = y=\frac{(x^{2}-5x + 6)}{(x-1)}$.

**Solution :**$y=\frac{(2x^{2}-4)}{(x-4)}$

To find y-intercept , plug in x= 0

0 =$y=\frac{(2x^{2}-4)}{(x-4)}$

So, $y=\frac{(2(0)^{2}-4)}{(0-4)}$

$y=\frac{-4}{-4}$

y=1

∴ y-intercepts is (0,1)

4) Find the y-intercept of the $f(x)=\frac{(x+2)}{(x+5)}$ rational function.

**Solution:**$f(x)=\frac{(x+2)}{(x+5)}$

To find y-intercept , plug in x= 0

$y=\frac{(0+2)}{(0 +5)}$

$y=\frac{2}{5}$

∴ y-intercepts are (0,$\frac{2}{5}$)

## Practice Questions

Q.1 Find x and y intercepts of the given rational function.1) $f(x)=\frac{(2x^{2}-1)}{(x-1)}$

2) $y =\frac{(x^{2}+ x -12)}{(x^{2}-9)}$

3) $f(x)=\frac{3}{(x-1)}$

Precalculus

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