Problem 48 - College Algebra

Question

For the following exercises, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph. $$ k(x)=\frac{2 x^{2}-3 x-20}{x-5} $$

Step-by-Step Solution

Answer

Vertical intercept at (0,4), horizontal intercepts at (4,0) and (-2.5,0), vertical asymptote at x=5, slant asymptote y=2x+7.

Step 1: Find the Vertical Intercept

To find the vertical intercept, set \(x = 0\) in the function \(k(x)\).\[k(x) = \frac{2(0)^{2} - 3(0) - 20}{0 - 5} = \frac{-20}{-5} = 4\]So, the vertical intercept is at the point \((0, 4)\).

Step 2: Find the Horizontal Intercept(s)

To find the horizontal intercepts, set \(k(x) = 0\). This occurs when the numerator equals zero:\[2x^{2} - 3x - 20 = 0\]Solving this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\):\[a = 2, \ b = -3, \ c = -20\]\[x = \frac{-(-3) \pm \sqrt{(-3)^2-4(2)(-20)}}{2(2)} = \frac{3 \pm \sqrt{9+160}}{4}\]\[x = \frac{3 \pm \sqrt{169}}{4} = \frac{3 \pm 13}{4}\]\[x = 4 \quad \text{or} \quad x = -2.5\]The horizontal intercepts are \((4,0)\) and \((-2.5,0)\).

Step 3: Identify the Vertical Asymptotes

Vertical asymptotes occur where the denominator equals zero and the numerator is not zero:\[x - 5 = 0 \ x = 5\]Since \(2(5)^2 - 3(5) - 20 eq 0\), there is a vertical asymptote at \(x = 5\).

Step 4: Determine the Horizontal or Slant Asymptote

With rational functions, the degree of the numerator is 2 (higher) and the degree of the denominator is 1 (lower), indicating a slant asymptote. Use polynomial long division to find it:Divide \(2x^{2} - 3x - 20\) by \(x - 5\):1. Divide the first terms: \(2x^2 \div x = 2x\).2. Multiply the entire divisor: \(2x(x-5) = 2x^2 - 10x\).3. Subtract: \((2x^2 - 3x) - (2x^2 - 10x) = 7x\).4. Bring down next term: \(7x - 20\).5. Divide as before: \(7x \div x = 7\).6. Multiply the entire divisor: \(7(x-5) = 7x - 35\).7. Subtract: \((7x - 20) - (7x - 35) = 15\).The slant asymptote is \(y = 2x + 7\).

Step 5: Sketch the Graph

Now that we have all the intercepts and asymptotes, we can sketch the graph:- **Vertical Intercept**: \((0, 4)\)- **Horizontal Intercepts**: \((4, 0)\) and \((-2.5, 0)\)- **Vertical Asymptote**: \(x = 5\)- **Slant Asymptote**: \(y = 2x + 7\)Mark these on a coordinate plane and draw the asymptotes. The graph will approach the asymptotes but never touch them and pass through the intercepts.

Key Concepts

Horizontal Intercept

When we talk about the horizontal intercepts of a function, we refer to the points where the graph crosses the x-axis. This means setting the function equal to zero and solving for the x-values that satisfy the equation.
For rational functions such as \(k(x) = \frac{2x^2 - 3x - 20}{x - 5}\), horizontal intercepts occur when the numerator equals zero.

• Step 1: Set the numerator equal to zero: \(2x^2 - 3x - 20 = 0\).
• Step 2: Solve the quadratic equation. In this case, using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) helps find the solutions \(x = 4\) and \(x = -2.5\).

Therefore, the horizontal intercepts are the points \(4, 0\) and \(-2.5, 0\). These points indicate where the graph crosses the x-axis, making them crucial for sketching the function.

Vertical Asymptote

Vertical asymptotes are the spots on a graph where the function approaches but never touches or crosses, usually caused by a zero in the denominator of a rational function.
To find the vertical asymptotes of \(k(x) = \frac{2x^2 - 3x - 20}{x - 5}\), identify the values of \(x\) that make the denominator zero:

• Step: Set the denominator \(x - 5 = 0\).
• Solution: Solve for \(x\), yielding \(x = 5\).

Check that the numerator at \(x = 5\) is not zero, which it isn't, confirming a vertical asymptote.
So, the vertical asymptote is at \(x = 5\), showing the line approaches but does not touch the graph as \(x\) approaches 5.

Slant Asymptote

Slant asymptotes, also known as oblique asymptotes, appear when the degree of the polynomial in the numerator is exactly one higher than the degree in the denominator in a rational function.
This occurs in our function \(k(x) = \frac{2x^2 - 3x - 20}{x - 5}\). To find it, perform polynomial long division:

• Divide \(2x^2\) by \(x\), giving \(2x\).
• Multiply \(2x\) by \(x - 5\), yielding \(2x^2 - 10x\).
• Subtract to get \(7x - 20\).
• Next, divide \(7x\) by \(x\), resulting in \(7\).
• Multiply \(7\) by \(x - 5\), providing \(7x - 35\).
• Subtract again, leaving \(15\), which is the remainder.

The divisor yields a slant asymptote equation, \(y = 2x + 7\), which is the line to which the graph of the function will closely approach as \(x\) moves to positive or negative infinity. This gives the graph a sort of diagonal direction.

Polynomial Long Division

Polynomial long division is a method that simplifies the division of one polynomial by another, similar to arithmetic long division. It is particularly useful for finding slant asymptotes.
In our example with \(k(x) = \frac{2x^2 - 3x - 20}{x - 5}\), polynomial long division helps us determine the equation of the slant asymptote:

• Start by dividing the first term of the numerator \(2x^2\) by the first term of the denominator \(x\), resulting in \(2x\).
• Next, multiply every term of the divisor \(x - 5\) by \(2x\). We obtain \(2x^2 - 10x\).
• Subtract this product from the polynomial in the dividend to find the next segment \(7x - 20\).
• Repeat the division process with \(7x\) divided by \(x\), giving you \(7\).
• Multiply \(7\) by each term in the divisor and subtract the result \(7x - 35\) from the current remainder \(7x - 20\), leaving a remainder of \(15\).

This step-by-step method constructs the slant asymptote equation \(y = 2x + 7\), simplifying the analysis of the rational function.