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1. which of the following is the graph of \\(y = -x^4 + 8x^2\\)?

Question

  1. which of the following is the graph of \\(y = -x^4 + 8x^2\\)?

Explanation:

Analyze the polynomial function

Using the End Behavior of Polynomials and Leading Coefficient Test knowledge points
The given function is:

$$ y = -x^4 + 8x^2 $$

This is a polynomial of degree \(n = 4\) (even) with a leading coefficient of \(a = -1\) (negative).
Since the degree is even and the leading coefficient is negative, the end behavior is:

$$ \text{As } x \to \infty, y \to -\infty \quad \text{and} \quad \text{As } x \to -\infty, y \to -\infty $$

Thus, both ends of the graph must point downwards.

Find the zeros and their multiplicities

Using the Multiplicity of Roots and Polynomial Graphing knowledge points
To find the \(x\)-intercepts, set \(y = 0\):

$$ -x^4 + 8x^2 = 0 \implies -x^2(x^2 - 8) = 0 $$

This yields the following roots:

  • \(x = 0\) with multiplicity 2 (even multiplicity, so the graph touches the \(x\)-axis and turns around at \(x = 0\)).
  • \(x = \pm\sqrt{8} \approx \pm 2.83\) with multiplicity 1 (odd multiplicity, so the graph crosses the \(x\)-axis at these points).

Match with the given graphs

Using the Polynomial Graphing knowledge point

  • The first graph has ends pointing in opposite directions (one up, one down), which is characteristic of an odd-degree polynomial.
  • The second graph has both ends pointing upwards, which corresponds to an even-degree polynomial with a positive leading coefficient (e.g., \(y = x^4 - 8x^2\)).
  • The third graph (partially visible at the bottom) shows a curve where both ends point downwards, touching the origin \((0,0)\) and turning around, which matches our analysis of \(y = -x^4 + 8x^2\).

Answer:

  • (A) Graph with one end pointing up and one end pointing down
  • (B) Graph with both ends pointing upwards
  • (C) Graph with both ends pointing downwards, touching the origin and turning around (Correct answer)