How do you graph f(x)=(2x^2-5x+2)/(2x^2-x-6) using holes, vertical and horizontal asymptotes, x and y intercepts?

Jul 18, 2018

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Explanation:

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The function:

color(red)(f(x)=(2x^2-5x+2)/(2x^2-x-6) Eqn.1 is rational.

color(green)("Step 1:"

Set $D R = 0$

$\Rightarrow 2 {x}^{2} - x - 6 = 0$

Factorize:

$\Rightarrow 2 {x}^{2} + 3 x - 4 x - 6 = 0$

$\Rightarrow x \left(2 x + 3\right) - 2 \left(2 x + 3\right) = 0$

$\Rightarrow \left(2 x + 3\right) \left(x - 2\right) = 0$

Factors of color(blue)(DR = (2x+3)(x-2) Res.1

color(green)("Step 2:"

Set $N R = 0$

$\Rightarrow 2 {x}^{2} - 5 x + 2 = 0$

Factorize:

$\Rightarrow 2 {x}^{2} - x - 4 x + 2 = 0$

$\Rightarrow x \left(2 x - 1\right) - 2 \left(2 x - 1\right) = 0$

$\Rightarrow \left(2 x - 1\right) \left(x - 2\right) = 0$

Hence, the factors of color(blue)(NR = (2x-1)(x-2) Res.2

color(green)("Step 3:"

Using the Res.1 and Res.2

Eqn.1 can now be rewritten as

color(blue)(f(x)= [(2x-1)(x-2)]/[(2x+3)(x-2)] Eqn.2

color(blue)(f(x)= [(2x-1)cancel((x-2))]/[(2x+3) cancel((x-2)]

color(blue)(f(x)=(2x-1)/(2x+3) Eqn.3

color(green)("Step 4:" Horizontal Asymptote

Refer to Eqn.1:

The leading coefficients of the highest terms are color(red)(2, 2

$\Rightarrow \frac{2}{2} = 1$

Horizontal Asymptote is at color(blue)(y=1

color(green)("Step 5:" Vertical Asymptote

Consider DR

$\Rightarrow 2 x + 3$

$\Rightarrow 2 x = - 3$

$\Rightarrow x = \left(- \frac{3}{2}\right)$

Vertical Asymptote is at color(blue)(x=(-3/2)

color(green)("Step 6:" Hole

Consider Eqn.2

color(blue)(f(x)= [(2x-1)(x-2)]/[(2x+3)(x-2)]

color(blue)((x-2) is the common factor,

Set color(blue)((x-2)=0

$\textcolor{b l u e}{x = 2}$

y-coordinate value of the hole

color(blue)(f(x)=(2x-1)/(2x+3) Eqn.3

Substitute color(red)(x=2

rArr (2*2 - 1)/(2*2+3

$\Rightarrow \frac{3}{7}$

Hence, the hole is at color(blue)((2, 3/7)

color(green)("Step 7:" x-intercept

Set color(blue)(2x-1=0

$\Rightarrow 2 x = 1$

$\Rightarrow x = \frac{1}{2}$

Hence, the x-intercept is at color(blue)((1/2, 0)

color(green)("Step 8:" y-intercept

Consider: $y = \frac{2 x - 1}{2 x + 3}$

Substitute color(red)(x=0

$\Rightarrow \frac{2 \cdot 0 - 1}{2 \cdot 0 + 3} = - \frac{1}{3}$

y-intercept is at color(red)((0, -1/3)

color(green)("Step 9:" Graph