F(x)=1/x24 Sketch graph

Set the denominator equal to zero:

This gives vertical asymptotes at $x = 2$ and $x = -2$.
There are no common factors in the numerator and denominator, so there are no holes.

The degree of the numerator (0) is less than the degree of the denominator (2).
Therefore, the horizontal asymptote is $y = 0$.

Set the numerator equal to zero:

This is impossible, so there are no x-intercepts.

Substitute $x = 0$ into the function:

The y-intercept is $\left(0, -\frac{1}{4}\right)$.

Substitute $-x$ into the function:

Since $F(-x) = F(x)$, the graph is symmetric about the y-axis.

We have vertical asymptotes at $x = -2$ and $x = 2$, and a horizontal asymptote at $y = 0$. The y-intercept is $(0, -1/4)$.
Let's pick points in the intervals $(-\infty, -2)$, $(-2, 2)$, and $(2, \infty)$:

• For $x = -3$: $F(-3) = \frac{1}{(-3)^2 - 4} = \frac{1}{9 - 4} = \frac{1}{5}$. Point: $\left(-3, \frac{1}{5}\right)$.
• For $x = -1$: $F(-1) = \frac{1}{(-1)^2 - 4} = \frac{1}{1 - 4} = -\frac{1}{3}$. Point: $\left(-1, -\frac{1}{3}\right)$.
• For $x = 1$: $F(1) = \frac{1}{(1)^2 - 4} = \frac{1}{1 - 4} = -\frac{1}{3}$. Point: $\left(1, -\frac{1}{3}\right)$.
• For $x = 3$: $F(3) = \frac{1}{(3)^2 - 4} = \frac{1}{9 - 4} = \frac{1}{5}$. Point: $\left(3, \frac{1}{5}\right)$.

1. Draw dashed vertical lines at $x = -2$ and $x = 2$.
2. Draw a dashed horizontal line at $y = 0$ (the x-axis).
3. Plot the y-intercept at $\left(0, -\frac{1}{4}\right)$.
4. Plot the additional points: $\left(-3, \frac{1}{5}\right)$, $\left(-1, -\frac{1}{3}\right)$, $\left(1, -\frac{1}{3}\right)$, $\left(3, \frac{1}{5}\right)$.
5. Connect the points, ensuring the graph approaches the asymptotes.
   • For $x < -2$, the graph is above the x-axis and approaches $y=0$ as $x \to -\infty$ and $x=-2$ as $x \to -2^-$.
   • For $-2 < x < 2$, the graph is below the x-axis, passing through $\left(0, -\frac{1}{4}\right)$, and approaches $x=-2$ as $x \to -2^+$ and $x=2$ as $x \to 2^-$.
   • For $x > 2$, the graph is above the x-axis and approaches $y=0$ as $x \to \infty$ and $x=2$ as $x \to 2^+$.

The key features for sketching the graph are:

• Vertical Asymptotes: $\boxed{x = -2, x = 2}$
• Horizontal Asymptote: $\boxed{y = 0}$
• X-intercepts: $\boxed{\text{None}}$
• Y-intercept: $\boxed{\left(0, -\frac{1}{4}\right)}$
• Symmetry: $\boxed{\text{About the y-axis}}$
• Additional points: $\boxed{\left(-3, \frac{1}{5}\right), \left(-1, -\frac{1}{3}\right), \left(1, -\frac{1}{3}\right), \left(3, \frac{1}{5}\right)}$