To determine where the graph of the function $h(x) = x^4 + 3x^2 - 4$ crosses the x-axis, we need to find the values of $x$ for which $h(x) = 0$.

Step 1: Set the Function Equal to Zero

$h(x) = x^4 + 3x^2 - 4 = 0$

Step 2: Substitute $y = x^2$

To simplify, let's let $y = x^2$, so $x^4 = y^2$. This transforms the equation into: $y^2 + 3y - 4 = 0$

Step 3: Factor the Quadratic Equation

Now, we factor $y^2 + 3y - 4$: $(y + 4)(y - 1) = 0$ This gives two solutions: $y = -4 \quad \text{and} \quad y = 1$

Since $y = x^2$ and $y = -4$ has no real solutions (since $x^2 \geq 0$), we discard it. So we only have: $y = 1$

Step 4: Solve for $x$

Since $y = x^2 = 1$, we solve for $x$: $x = \pm 1$

Conclusion

The graph of $h(x) = x^4 + 3x^2 - 4$ crosses the x-axis at $x = -1$ and $x = 1$.

Answer

The correct answer is:

• Option C: $x = -1$ and $x = 1$

Would you like further details on this solution, or do you have any other questions?

Here are some related questions:

1. How do we determine the x-intercepts of polynomial functions in general?
2. What other methods can we use to solve equations with $x^4$ terms?
3. How do transformations of quadratic forms help in factoring higher-degree polynomials?
4. Why can we disregard the solution $y = -4$ in this problem?
5. What does it mean graphically when a function crosses versus touches the x-axis?

Tip: When dealing with higher-degree polynomials, using substitutions can often simplify the equation and make it easier to solve.