Problem 23 - College Algebra

Question

For the following exercises, find the \(x\) - and \(y\) -intercepts for the functions. $$ f(x)=\frac{x^{2}+x+6}{x^{2}-10 x+24} $$

Step-by-Step Solution

Answer

y-intercept: (0, 1/4); no real x-intercepts.

Step 1: Find the y-intercept

To find the y-intercept of the function, substitute \(x = 0\) into the function. This gives us: \[f(0) = \frac{0^{2} + 0 + 6}{0^{2} - 10 \times 0 + 24} = \frac{6}{24} = \frac{1}{4}\] Hence, the y-intercept is \((0, \frac{1}{4})\).

Step 2: Set the numerator equal to zero for x-intercepts

The x-intercept(s) of a function occur when the function equals zero. For a rational function, this means setting the numerator equal to zero and solving for \(x\). The numerator is \(x^2 + x + 6\). Setting this to zero gives us:\[x^2 + x + 6 = 0\]We'll solve this quadratic equation for \(x\) using the quadratic formula or by factoring if possible. Factoring doesn't work here, so use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 1\), \(b = 1\), and \(c = 6\).

Step 3: Apply the quadratic formula

Calculate using the quadratic formula with \(a = 1\), \(b = 1\), and \(c = 6\):\[x = \frac{-1 \pm \sqrt{1^2 - 4(1)(6)}}{2(1)} = \frac{-1 \pm \sqrt{1 - 24}}{2}\]\[x = \frac{-1 \pm \sqrt{-23}}{2}\]Since the discriminant \(-23\) is negative, the result involves imaginary numbers. Therefore, there are no real x-intercepts.

Step 4: Verify real x-intercepts don't exist

The absence of a real, positive discriminant suggests there are no real solutions to \(x^2 + x + 6 = 0\). This means the function has no x-intercepts.

Key Concepts

Intercepts

Intercepts are points where a graph crosses the axes. For functions, we typically seek two types of intercepts:

• **x-intercepts**: these are points where the graph crosses the x-axis, meaning the function's output (y-value) is zero.
• **y-intercepts**: these are points where the graph crosses the y-axis, which occur at the input value of zero for x.

To find the y-intercept of a function, substitute 0 for all instances of x in the equation and solve for f(x). For the function given, substituting 0 results in the calculation: \[ f(0) = \frac{0^2 + 0 + 6}{0^2 - 10 \times 0 + 24} = \frac{6}{24} = \frac{1}{4} \]This means the y-intercept is at point \( (0, \frac{1}{4}) \).For x-intercepts, set the numerator of the fraction to zero and solve for x; this reveals where the dependent variable, y, reaches zero. In this exercise, when you try this, you find the x-intercepts by setting \( x^2 + x + 6 = 0 \), which leads us to the Quadratic Formula discussion since direct factoring is not possible.

Quadratic Formula

The Quadratic Formula offers a method to solve any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula gives up to two solutions for x, referred to as the roots of the equation. However, the discriminant, which is the \( b^2 - 4ac \) part under the square root, tells us about the nature of these solutions.

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is one real root (a repeated root).
• If negative, the roots are complex or imaginary, meaning the equation has no real solutions.

In the provided exercise, the equation \( x^2 + x + 6 = 0 \) results in a discriminant of \(-23\), indicating no real solutions. Hence, there are no x-intercepts for the function.

Rational Functions

Rational functions are ratios of polynomials, of the form \( f(x) = \frac{p(x)}{q(x)} \), where both \( p(x) \) and \( q(x) \) are polynomials. These functions are interesting because they can have vertical asymptotes, horizontal asymptotes, and intercepts, just like the one we're analyzing.

• **Vertical asymptotes** typically occur where the denominator \( q(x) \) equals zero, provided the numerator \( p(x) \) doesn't simultaneously equal zero (which would be a hole instead).
• **Horizontal asymptotes** give an idea of how the function behaves as \( x \) approaches infinity or negative infinity.

Finding intercepts, as we've done, involves examining the points where the function equals zero (x-intercepts) or crosses the y-axis (y-intercepts). Knowing these properties helps to graph the function and predict its behavior across its domain. In the exercise, we've explored this by finding the y-intercept and attempting to find x-intercepts using the Quadratic Formula.