(4x+4)÷(4x^3-14x^2+18x)

Question

Whats the HA ,VA,RD,x-intercept and y intercept

Answer 1

The first thing we need to do is factor out the denominator We need to see if any of the factors cancel with the numerator and figure out the VA:

4x³ - 14x² + 18x

x(4x² - 14x + 18)

2x(2x² - 7x + 9)

 

Let's use the quadratic formula to see if we can factor any further:

(-b ± √(b² - 4ac)) / 2a

(7 ± √(7² - 4*2*9) / (2 * 2)

(7 ± √(49 - 72) / 4

(7 ± √-23) / 4

We can't take the square root of a negative, so we're done there.

 

We can factor then numerator as well:

4x + 4

4(x + 1)

 

So now we check for cancellations:

4(x + 1) / 2x(2x² - 7x + 9)

2(x + 1) / x(2x² - 7x + 9)

 

To find the HA, we take a look at the highest powers and Coefficients:

4x / 4x³

Since the denominator has a higher power, the HA is 0

 

To find the VA, we set the denominator to 0:

x(2x² - 7x + 9) = 0

x = 0, and as we found out with the quadratic formula, 2x² - 7x + 9 is never 0, so the VA is 0.

 

To find the x-intercept, we set the numerator to 0:

2(x + 1) = 0

x + 1 = 0

x = -1

 

To find the y-intercept, we set x to 0, but we can save ourselves some trouble--we discovered the VA at 0, so it will never actually cross the y-intercept.

Answer 2

We can start by factoring the equation.

 

(4(x + 1)) / (2x(2x² - 7x + 9))

 

To find the VA, we set the denominator to zero.

 

2x(2x² - 7x + 9) = 0

 

2x = 0      and      2x² - 7x + 9 = 0

  x = 0

 

 

We can find the other value of x by using the quadratic formula.

 

x = (-b ± √(b² - 4ac)) / 2a

 

where:

a = 2

b = -7

c = 9

 

x = (7 ± √(49 - 4(18))) / 4

x = (7 ± √(-23)) / 4

x = (7 ± i√(23)) / 4

 

x = (7 + i√(23)) / 4        and      x = (7 - i√(23)) / 4

 

 

VA = 0 ,  (7 + i√(23)) / 4   ,   and  (7 - i√(23)) / 4

 

 

To find x-intercept, set the numerator of the equation to zero.

 

4(x + 1) = 0

 

x + 1 = 0

x = -1

 

x-intercept is (-1,0)

 

 

To find y-intercept, evaluate the equation when x=0.  In this case, there is no y-intercept because one of the VA's is zero, which means the graph will never cross the y-axis.

 

To find the HA, we look at the degree of the equation.  The numerator has a degree of 1 because the exponent of the leading term is 1.  The denominator has a degree of 3 because the exponent of the leading term is 3.  When the degree in the numerator is less than the degree in the denominator, the HA is 0.

 

 

If you are interested in learning another method to find the HA, let me know.  You will learn this method when you reach calculus.