1. The problem is to graph the function $f(x) = 5x^4 - 13$. 2. This is a polynomial function of degree 4, which means it is a quartic function. 3. To understand the graph, let's analyze key features: - The leading term is $5x^4$, which dominates for large $|x|$ and is always positive, so the graph will rise to infinity as $x \to \pm \infty$. - The constant term is $-13$, which shifts the graph down by 13 units. 4. Find the y-intercept by evaluating $f(0)$: $$f(0) = 5 \cdot 0^4 - 13 = -13$$ So the graph crosses the y-axis at $(0, -13)$. 5. Find critical points by taking the derivative: $$f'(x) = 20x^3$$ Set $f'(x) = 0$ to find critical points: $$20x^3 = 0 \implies x = 0$$ 6. Determine the nature of the critical point at $x=0$ by the second derivative: $$f''(x) = 60x^2$$ At $x=0$, $f''(0) = 0$, so the test is inconclusive. However, since $f(x)$ is even and quartic, $x=0$ is a minimum point. 7. Evaluate $f(0) = -13$ confirms the minimum value. 8. The graph is symmetric about the y-axis because the function is even. Final answer: The graph of $f(x) = 5x^4 - 13$ is a quartic curve opening upwards with a minimum at $(0, -13)$ and y-intercept at $(0, -13)$.