Question

y=2x^3-5x^2-3x
how many turns are there


dont get this wrong

Answer 1

To determine the number of turns in the function \( y = 2x^3 - 5x^2 - 3x \), we need to find its critical points by calculating the derivative and then determining where it changes direction.

1. **Find the derivative** of \( y \):
\[
y' = \frac{d}{dx}(2x^3 - 5x^2 - 3x) = 6x^2 - 10x - 3
\]

2. **Set the derivative equal to zero** to find critical points:
\[
6x^2 - 10x - 3 = 0
\]

3. **Solve the quadratic equation** using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 6 \), \( b = -10 \), and \( c = -3 \):
\[
x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 6 \cdot (-3)}}{2 \cdot 6}
\]
\[
= \frac{10 \pm \sqrt{100 + 72}}{12}
\]
\[
= \frac{10 \pm \sqrt{172}}{12}
\]
\[
= \frac{10 \pm 2\sqrt{43}}{12} = \frac{5 \pm \sqrt{43}}{6}
\]

4. **Calculate the discriminant** to confirm the number of solutions:
The discriminant \( b^2 - 4ac = 172 \), which is positive. This indicates that there are two distinct real roots.

Since there are two distinct critical points, the number of turns (local maxima or minima) in the function is **two**.