Question

If the function \(y=\frac{x^{2}+b x+c}{4 x^{2}+29 x+c},\) where \(b\) and \(c\) are real numbers, has a point of discontinuity at \(\left(-8, \frac{11}{35}\right),\) where does it have \(x\) -intercept(s) and vertical asymptote(s), if any?

Short answer

The function has x-intercepts at \(x = \frac{-b \pm \sqrt{b^2 + 96}}{2}\). The vertical asymptotes are at \( x = \frac{3}{4} \) and \( x = -8 \).

Step-by-step solution

Identify the Point of Discontinuity

The function has a point of discontinuity at \((-8, \frac{11}{35})\). This implies that the denominator \(4x^2 + 29x + c\) must be zero when \(x = -8\).

Determine the Value of c

Substitute \(x = -8\) into the denominator to get: \[ 4(-8)^2 + 29(-8) + c = 0. \] This simplifies to \[ 256 - 232 + c = 0 \Rightarrow c = -24. \]

Using the Point of Discontinuity in the Numerator

Plug \(x = -8\) and \(c = -24\) into the numerator to check: \[ \frac{256 - 8b - 24}{0} = \frac{11}{35}. \] This equation is not used further since it confirms the discontinuity.

Find x-intercepts

Set the numerator equal to zero: \[x^2 + bx - 24 = 0. \] Solve this quadratic equation to find \(x\)-intercepts.

Solve the Quadratic Equation

The roots are the \(x\)-intercepts. Since this is a quadratic \[x^2 + bx - 24 = 0, \] the solutions can be found using the quadratic formula: \ x = \frac{-b \pm \sqrt{b^2 + 96}}{2} \.

Find Vertical Asymptotes

Set the denominator equal to zero to solve for vertical asymptotes: \[4x^2 + 29x - 24 = 0. \] Solve this quadratic equation using the quadratic formula: \ x = \frac{-29 \pm \sqrt{841 + 384}}{8} = \frac{-29 \pm 35}{8}. \

Simplify the Solutions for Asymptotes

The solutions are: \[x = \frac{6}{8} = \frac{3}{4} \] and \[ x = \frac{-64}{8} = -8. \]

Key concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It can find the roots of any quadratic equation. The formula is:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here, \(a\), \(b\), and \(c\) are coefficients from the quadratic equation. The \(\pm\) symbol indicates that there will be two solutions, commonly called roots.

In the given exercise, we encountered the quadratic formula twice, once to find the x-intercepts and once to find the vertical asymptotes. The quadratic formula allows us to solve for \(x\) by substituting the coefficients into the formula and simplifying it.

• For finding the x-intercepts, the quadratic equation was \(x^2 + bx - 24 = 0\).
• For finding the vertical asymptotes, the quadratic equation was \(4x^2 + 29x - 24 = 0\).
  

Once we have these quadratic equations, we can use the quadratic formula by plugging in the values of \(a\), \(b\), and \(c\) to find the roots or the solutions. The discriminant \(b^2 - 4ac\) helps in determining the nature of the roots—whether they are real or complex.

Vertical Asymptotes

Vertical asymptotes are vertical lines where the function approaches infinity. They occur in rational functions where the denominator goes to zero. To find these asymptotes, we set the denominator equal to zero and solve for \(x\).

In our exercise, the denominator was \(4x^2 + 29x - 24\). To find the vertical asymptotes, we set this equal to zero and solved the quadratic equation:
\[4x^2 + 29x - 24 = 0\]
Using the quadratic formula, we find:
\[x = \frac{-29 \pm \sqrt{841 + 384}}{8}\]
This simplifies to:
\[x = \frac{-29 \pm 35}{8}\]
Which gives us the solutions \(\frac{3}{4}\) and \(-8\).

These are the vertical asymptotes for the given function, indicating where the function will shoot off to positive or negative infinity.

X-Intercepts

X-intercepts are points where the graph of a function crosses the x-axis. At these points, the value of y is zero. To find the x-intercepts of a rational function, we set the numerator equal to zero.

In the problem, the numerator is \(x^2 + bx - 24\). Setting this to zero and solving the quadratic equation will give the x-intercepts:
\[x^2 + bx - 24 = 0\]
We use the quadratic formula again:
\[x = \frac{-b \pm \sqrt{b^2 + 96}}{2}\]
Here, we solve for the values of \(x\) that make the numerator zero, giving us the x-intercepts.

Finding x-intercepts is crucial because it tells us where the function touches the x-axis, and these points are key in understanding the overall behavior and graph of the function.