Question

(x) = \frac{2x^2 - 1x - 3}{2x^2 - 5x - 12}
unction has:
intercept at the point
tercepts at the point(s)
ical asymptotes at x =
zontal asymptote at y =
n 7
zeros of the function.
4x^3 - 20x^2 + 20x - 16
enter the zeros (exact values, not decimal ap

Explanation:

Part 1: Analyzing the Rational Function \( f(x) = \frac{2x^2 - x - 3}{2x^2 - 5x - 12} \)

1. y - intercept

To find the y - intercept, we set \( x = 0 \) in the function.

Step 1: Substitute \( x = 0 \) into the function

\( f(0)=\frac{2(0)^2-0 - 3}{2(0)^2-5(0)-12}=\frac{- 3}{-12}=\frac{1}{4} \)
The y - intercept is at the point \( (0,\frac{1}{4}) \)

2. x - intercepts

To find the x - intercepts, we set the numerator equal to zero (and ensure the denominator is not zero at those points)

Step 1: Solve \( 2x^2-x - 3=0 \)

We factor the quadratic equation: \( 2x^2-x - 3=(2x - 3)(x + 1)=0 \)
Setting each factor equal to zero:

• \( 2x-3 = 0\Rightarrow x=\frac{3}{2} \)
• \( x + 1=0\Rightarrow x=-1 \)

We check the denominator at these values:

• When \( x=\frac{3}{2} \), denominator \( 2(\frac{3}{2})^2-5(\frac{3}{2})-12=2\times\frac{9}{4}-\frac{15}{2}-12=\frac{9}{2}-\frac{15}{2}-12=- 3 - 12=-15

eq0 \)

• When \( x = - 1 \), denominator \( 2(-1)^2-5(-1)-12=2 + 5-12=-5

eq0 \)

So the x - intercepts are at the points \( (-1,0) \) and \( (\frac{3}{2},0) \)

3. Vertical Asymptotes

To find the vertical asymptotes, we set the denominator equal to zero and solve for \( x \)

Step 1: Solve \( 2x^2-5x - 12 = 0 \)

Factor the quadratic: \( 2x^2-5x - 12=(2x + 3)(x - 4)=0 \)
Setting each factor equal to zero:

• \( 2x+3=0\Rightarrow x=-\frac{3}{2} \)
• \( x - 4=0\Rightarrow x = 4 \)

4. Horizontal Asymptote

For a rational function \( f(x)=\frac{ax^n+...}{bx^m+...} \), if \( n = m \), the horizontal asymptote is \( y=\frac{a}{b} \)
Here, the degree of the numerator \( n = 2 \) and the degree of the denominator \( m = 2 \), and \( a = 2 \), \( b = 2 \)
So the horizontal asymptote is \( y=\frac{2}{2}=1 \)

Part 2: Finding the zeros of \( 4x^3-20x^2 + 20x-16 \)

First, we can factor out a common factor of 4:

Step 1: Factor out 4

\( 4x^3-20x^2 + 20x-16=4(x^3-5x^2 + 5x - 4) \)

We try to find a root of the cubic equation \( x^3-5x^2 + 5x - 4 = 0 \) by trial and error. We test \( x = 4 \):

Step 2: Test \( x = 4 \) in \( x^3-5x^2 + 5x - 4 \)

\( 4^3-5\times4^2+5\times4 - 4=64-80 + 20 - 4=0 \)
So \( (x - 4) \) is a factor of \( x^3-5x^2 + 5x - 4 \)

We perform polynomial long - division or use synthetic division to divide \( x^3-5x^2 + 5x - 4 \) by \( (x - 4) \)

Using synthetic division:
The coefficients of the cubic are \( 1,-5,5,-4 \) and we divide by \( x - 4 \)

4 | 1  -5  5  -4
        4  -4   4
      1  -1  1   0

So \( x^3-5x^2 + 5x - 4=(x - 4)(x^2-x + 1) \)

The quadratic \( x^2-x + 1 \) has discriminant \( \Delta=(-1)^2-4\times1\times1=1 - 4=-3<0 \), so it has no real roots.

So the zero of the function \( 4x^3-20x^2 + 20x - 16 \) is \( x = 4 \) (since the other factor \( x^2-x + 1 \) has no real zeros)

Answer:

s

For the rational function:

• y - intercept: \( (0,\frac{1}{4}) \)
• x - intercepts: \( (-1,0),(\frac{3}{2},0) \)
• Vertical asymptotes: \( x = -\frac{3}{2},x = 4 \)
• Horizontal asymptote: \( y = 1 \)

For the cubic function \( 4x^3-20x^2 + 20x - 16 \):

The zero is \( x = 4 \)