Functions & Graphs - Free Printable

Problem: Analyze the cubic function \( y = x^3 - 4x^2 + 3 \)

We are tasked with:
1. Drawing the graph of \( y = x^3 - 4x^2 + 3 \).
2. Estimating the roots of the equation \( y = x^3 - 4x^2 + 3 \).
3. Estimating the coordinates of the turning points of the graph.

#### Step 1: Calculate the values of \( y \) for given \( x \)-values
We are provided with a table of \( x \)-values: \( x = -2, -1, 0, 1, 2, 3, 4 \). We will calculate the corresponding \( y \)-values using the function \( y = x^3 - 4x^2 + 3 \).

- For \( x = -2 \):
\[
y = (-2)^3 - 4(-2)^2 + 3 = -8 - 4(4) + 3 = -8 - 16 + 3 = -21
\]

- For \( x = -1 \):
\[
y = (-1)^3 - 4(-1)^2 + 3 = -1 - 4(1) + 3 = -1 - 4 + 3 = -2
\]

- For \( x = 0 \):
\[
y = (0)^3 - 4(0)^2 + 3 = 0 - 0 + 3 = 3
\]

- For \( x = 1 \):
\[
y = (1)^3 - 4(1)^2 + 3 = 1 - 4(1) + 3 = 1 - 4 + 3 = 0
\]

- For \( x = 2 \):
\[
y = (2)^3 - 4(2)^2 + 3 = 8 - 4(4) + 3 = 8 - 16 + 3 = -5
\]

- For \( x = 3 \):
\[
y = (3)^3 - 4(3)^2 + 3 = 27 - 4(9) + 3 = 27 - 36 + 3 = -6
\]

- For \( x = 4 \):
\[
y = (4)^3 - 4(4)^2 + 3 = 64 - 4(16) + 3 = 64 - 64 + 3 = 3
\]

The completed table is:
\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\
\hline
y & -21 & -2 & 3 & 0 & -5 & -6 & 3 \\
\hline
\end{array}
\]

#### Step 2: Draw the graph of \( y = x^3 - 4x^2 + 3 \)
Using the calculated points \((-2, -21)\), \((-1, -2)\), \((0, 3)\), \((1, 0)\), \((2, -5)\), \((3, -6)\), and \((4, 3)\), we can plot these points on the graph and sketch the curve. The graph of a cubic function generally has a "S"-shape or a "wave" shape, depending on its coefficients.

#### Step 3: Estimate the roots of \( y = x^3 - 4x^2 + 3 \)
The roots of the equation are the \( x \)-values where \( y = 0 \). From the table, we see that \( y = 0 \) when \( x = 1 \). To find the other roots, we can use the fact that a cubic polynomial has at most three real roots. By examining the graph, we can see that there are two more roots, one between \( x = -1 \) and \( x = 0 \), and another between \( x = 2 \) and \( x = 3 \). Using numerical methods or a graphing calculator, we can approximate these roots. However, for this problem, we can estimate them visually from the graph.

From the graph, the approximate roots are:
- \( x \approx -0.5 \)
- \( x = 1 \)
- \( x \approx 3.5 \)

#### Step 4: Estimate the coordinates of the turning points
Turning points occur where the derivative of the function is zero. The derivative of \( y = x^3 - 4x^2 + 3 \) is:
\[
y' = 3x^2 - 8x
\]
Setting \( y' = 0 \):
\[
3x^2 - 8x = 0 \implies x(3x - 8) = 0 \implies x = 0 \text{ or } x = \frac{8}{3}
\]
So, the turning points occur at \( x = 0 \) and \( x = \frac{8}{3} \approx 2.67 \).

- For \( x = 0 \):
\[
y = 0^3 - 4(0)^2 + 3 = 3
\]
So, the turning point is \( (0, 3) \).

- For \( x = \frac{8}{3} \):
\[
y = \left(\frac{8}{3}\right)^3 - 4\left(\frac{8}{3}\right)^2 + 3 = \frac{512}{27} - 4\left(\frac{64}{9}\right) + 3 = \frac{512}{27} - \frac{256}{9} + 3 = \frac{512}{27} - \frac{768}{27} + \frac{81}{27} = \frac{512 - 768 + 81}{27} = \frac{-175}{27} \approx -6.48
\]
So, the turning point is approximately \( \left(\frac{8}{3}, -6.48\right) \).

Final Answers:

1. The graph of \( y = x^3 - 4x^2 + 3 \) is sketched using the points from the table.
2. The estimated roots are \( x \approx -0.5 \), \( x = 1 \), and \( x \approx 3.5 \).
3. The estimated coordinates of the turning points are \( (0, 3) \) and \( \left(\frac{8}{3}, -6.48\right) \).

\[
\boxed{x \approx -0.5, x = 1, x \approx 3.5, (0, 3), \left(\frac{8}{3}, -6.48\right)}
\]