Question

Graph the function y = x3 + 3x2 – x – 3. Which lists all of the turning points of the graph rounded to the nearest whole number?

(–3, 0) and (1, 0)
(–2, 3) and (0, –3)
(–2, 3), (–1, 0), and (0, –3)
(–3, 0), (–1, 0), and (1,0)

Answer 1

To find the turning points of the function \( y = x^3 + 3x^2 - x - 3 \), we first need to compute the first derivative and set it to zero. This will help us locate where the slope of the curve is zero (i.e., potential turning points).

1. **Finding the first derivative**:
\[
\frac{dy}{dx} = 3x^2 + 6x - 1
\]

2. **Setting the first derivative equal to zero**:
\[
3x^2 + 6x - 1 = 0
\]

To solve this quadratic equation, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 3, b = 6, c = -1 \).

Calculate the discriminant:
\[
b^2 - 4ac = 6^2 - 4(3)(-1) = 36 + 12 = 48
\]

Now, plug in the values:
\[
x = \frac{-6 \pm \sqrt{48}}{2 \cdot 3} = \frac{-6 \pm 4\sqrt{3}}{6} = \frac{-3 \pm \frac{2\sqrt{3}}{3}}{3} = \frac{-3 \pm \frac{2\sqrt{3}}{3}}{3}
\]

Approximating the roots:
\[
x_1 = -2 - \frac{2\sqrt{3}}{3} \approx -3.15 \quad \text{(rounded to -3)}
\]
\[
x_2 = -2 + \frac{2\sqrt{3}}{3} \approx -0.15 \quad \text{(rounded to -1)}
\]

3. **Finding the corresponding y-values**:
Now substitute \( x_1 \) and \( x_2 \) back into the original function \( y(x) \) to find the corresponding y-values at the turning points:

For \( x = -2 \):
\[
y(-2) = (-2)^3 + 3(-2)^2 - (-2) - 3 = -8 + 12 + 2 - 3 = 3 \quad \text{(point (-2, 3))}
\]

For \( x = -1 \):
\[
y(-1) = (-1)^3 + 3(-1)^2 - (-1) - 3 = -1 + 3 + 1 - 3 = 0 \quad \text{(point (-1, 0))}
\]

For \( x = 0 \):
\[
y(0) = (0)^3 + 3(0)^2 - (0) - 3 = -3 \quad \text{(point (0, -3))}
\]

4. **List of turning points**:
The turning points of the function, rounded to the nearest whole number, are:
- \((-2, 3)\)
- \((-1, 0)\)
- \((0, -3)\)

So the correct list of turning points is:
**(–2, 3), (–1, 0), and (0, –3)**.