Question

follow the seven step strategy to graph the following rational function.
$f(x)=\frac{3x^2 + x - 4}{2x^2 - 7x}$

origin symmetry
y-axis symmetry
neither y-axis symmetry nor origin symmetry
what is the y-intercept? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the y-intercept is

b. there is no y-intercept.
what is/are the x-intercept(s)? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a.
the x-intercept(s) is/are $1, -\frac{4}{3}$.
(type an integer or a simplified fraction. use a comma to separate answers as needed.)
b. there is no x-intercept.
find the vertical asymptote(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the equation(s) of the vertical asymptote(s) is/are
(type an equation. use a comma to separate answers as needed.)
b. there is no vertical asymptote.

Explanation:

Step1: Find the domain

For the function \(f(x)=\frac{3x^{2}+x - 4}{2x^{2}-7x}\), the denominator \(2x^{2}-7x=x(2x - 7)\). Set \(x(2x - 7)=0\), we get \(x = 0\) or \(x=\frac{7}{2}\). The domain is \(\mathbb{R}-\{0,\frac{7}{2}\}\).

Step2: Check for symmetry

• Origin symmetry: Check if \(f(-x)=-f(x)\).

\(f(-x)=\frac{3(-x)^{2}+(-x)-4}{2(-x)^{2}-7(-x)}=\frac{3x^{2}-x - 4}{2x^{2}+7x}\), and \(-f(x)=-\frac{3x^{2}+x - 4}{2x^{2}-7x}=\frac{-3x^{2}-x + 4}{2x^{2}-7x}\). Since \(f(-x)
eq -f(x)\).

• y - axis symmetry: Check if \(f(-x)=f(x)\). Since \(f(-x)=\frac{3x^{2}-x - 4}{2x^{2}+7x}

eq f(x)=\frac{3x^{2}+x - 4}{2x^{2}-7x}\). So, the function has neither \(y\) - axis symmetry nor origin symmetry.

Step3: Find the \(y\) - intercept

The \(y\) - intercept is found by setting \(x = 0\). But \(x = 0\) is not in the domain of \(f(x)\) (because the denominator is \(0\) when \(x = 0\)). So, there is no \(y\) - intercept.

Step4: Find the \(x\) - intercept

Set the numerator equal to \(0\), i.e., \(3x^{2}+x - 4=0\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 3\), \(b = 1\), \(c=-4\)).
\(x=\frac{-1\pm\sqrt{1^{2}-4\times3\times(-4)}}{2\times3}=\frac{-1\pm\sqrt{1 + 48}}{6}=\frac{-1\pm7}{6}\).
We get \(x = 1\) or \(x=-\frac{4}{3}\).

Step5: Find the vertical asymptote

Set the denominator equal to \(0\) (after factoring and canceling any common factors with the numerator). Since \(2x^{2}-7x=x(2x - 7)\) and there are no common factors with the numerator \(3x^{2}+x - 4=(3x + 4)(x - 1)\).
Set \(x(2x - 7)=0\), we get \(x = 0\) and \(x=\frac{7}{2}\) as the equations of the vertical asymptotes.

Answer:

• Symmetry: neither \(y\) - axis symmetry nor origin symmetry.
• \(y\) - intercept: B. There is no \(y\) - intercept.
• \(x\) - intercept: A. The \(x\) - intercept(s) is/are \(1,-\frac{4}{3}\).
• Vertical asymptote: A. The equation(s) of the vertical asymptote(s) is/are \(x = 0,x=\frac{7}{2}\).