1. Let's start by understanding the given function. You mentioned it's $y=4x^2$ but with a $+x$ on top of the $4$. This suggests the function is $y=\frac{x+4}{x^2}$. 2. Rewrite the function clearly as $$y=\frac{x+4}{x^2}$$ 3. To analyze this function, break it into two parts: $$y=\frac{x}{x^2} + \frac{4}{x^2} = \frac{1}{x} + \frac{4}{x^2}$$ 4. This function is defined for all real $x$ except $x=0$ where the denominator is zero. 5. You can check intercepts: - No $y$-intercept since $x=0$ is undefined. - $x$-intercept at $y=0$, so solve $\frac{x+4}{x^2} = 0$ which implies $x+4=0$, so $x=-4$. 6. The function decreases rapidly near $x=0$ and approaches $0$ as $x \to \pm \infty$. Therefore, the function is $$y=\frac{x+4}{x^2}$$ with the key features described above.