# What are the x intercepts for the rational function y = (2x2-10x-6)/ (2x2-x-3).

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### 3 Answers

The x-intercepts are nothing but the roots of the function, as at the roots the value of the function becomes 0.

Now we have y = (2x^2 - 10x - 6)/ (2x^2 - x - 3).

We have to find the values of x for which y = (2x^2 - 10x - 6)/ (2x^2 - x - 3) = 0

=> (2x^2 - 10x - 6) = 0

x1 = [-b + sqrt (b^2 - 4ac)]/2a

=> [ 10 + sqrt ( 100 + 48)]/4

=> 2.5 + (sqrt 37)/2

x2 = 2.5 - (sqrt 37)/2

**Therefore the x- intercepts are at **

**(2.5 - (sqrt 37)/2, 0) and ( 2.5 + (sqrt 37)/2 , 0)**

y = (2x^2 - 10x - 6) / (2x^2 - x - 3)

We need to determine the x-intercept.

x-intercept is when the curve meets the x-axis.

Then, y value will be 0.

Then we need to find the x-coordinate.

==> 0 = (2x^2 - 10x -6)/(2x^2 - x -3)

==> 2x^2 - 10x - 6 = 0

Then we will divide by 2:

==> x^2 - 5x - 3 = 0

Now we will solve for x values using the formula.

==> x1= ( 5 + sqrt(25+4*3) / 2 = (5+ sqrt(37) / 2

==> x2= ( 5- sqrt37) / 2

Then we have two intersections points between the curve and the x-axis.

**==> x-intercepts are: ( 5+sqrt37) /2 , 0) and ( (5-sqrt37)/2 , 0)**

x intercept means same as where the x axis cuts the curve represented by rational function y = 2x2-10x-6/ (2x2-x-3). The equation of the x axis is y= 0. So we put y= 0 in y = 2x2-10x-6/( 2x2-x-3) and for x.

=> 0 = 2x2-10x-6/( 2x2-x-3).

Multiply both sides by (2x^2-x-3) and we get:

0 = 2x^2-10x-6 .

We divide by 2 and interchange sides:

x^2-5x-3 = 0. This is quadratic equation of the type ax^2+bx+c = 0 whose roots are x1 = {-b+(b^2-4ac)^(1/2)}/2a, or x2 = {-b-(b^2-4ac)^(1/2)}/2a.

So the the roots of x^2-5x-3 = 0 are x1 = {5+(5^2-4*1*-3)^(1/2)}/2 = {5+37^(1/2)}/2 or x2 = {5+37^(1/2)}/2.

Therefore the required x intercepts are x1 = {5+37^(1/2)}/2 and x2 = {5-37^(1/2)}/2.