Question

Use the graph of the polynomial equation
y=x^4-10x^2+16 to describe its concavity. How many points of inflection does this curve have?

Answer 1

To analyze the concavity of the polynomial function \( y = x^4 - 10x^2 + 16 \), we first need to find its second derivative. The steps are as follows:

1. **First Derivative**:
\[
y' = \frac{d}{dx}(x^4 - 10x^2 + 16) = 4x^3 - 20x
\]

2. **Second Derivative**:
\[
y'' = \frac{d}{dx}(4x^3 - 20x) = 12x^2 - 20
\]

3. **Finding Points of Inflection**:
Points of inflection occur where the second derivative is equal to zero or does not exist. We set the second derivative equal to zero:
\[
12x^2 - 20 = 0
\]
Rearranging gives:
\[
12x^2 = 20 \quad \Rightarrow \quad x^2 = \frac{20}{12} = \frac{5}{3}
\]
Taking the square root of both sides, we find:
\[
x = \pm \sqrt{\frac{5}{3}} = \pm \frac{\sqrt{15}}{3}
\]

4. **Number of Points of Inflection**:
Since we found two distinct values of \( x \) ( \( x = \frac{\sqrt{15}}{3} \) and \( x = -\frac{\sqrt{15}}{3} \) ), there are **two points of inflection**.

5. **Concavity Analysis**:
To determine the concavity, we examine the sign of the second derivative \( y'' = 12x^2 - 20 \):
- For \( x < -\sqrt{\frac{5}{3}} \), \( y'' > 0 \) (the polynomial is concave up).
- For \( -\sqrt{\frac{5}{3}} < x < \sqrt{\frac{5}{3}} \), \( y'' < 0 \) (the polynomial is concave down).
- For \( x > \sqrt{\frac{5}{3}} \), \( y'' > 0 \) (the polynomial is again concave up).

Thus, the polynomial exhibits the following:
- Concave up on intervals \( (-\infty, -\sqrt{\frac{5}{3}}) \) and \( (\sqrt{\frac{5}{3}}, \infty) \).
- Concave down on the interval \( (-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}) \).

In summary, the curve has **two points of inflection**, and its concavity changes from concave up to concave down and back to concave up at these points.