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Sketch the graph of the function y = x3 − 2x2 − 3x. Find the intercepts and turning points.

Mathematics

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Graph of y = x³ − 2x² − 3x

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Key points of y = x³ − 2x² − 3x
Pointxy
x-intercept−10
Maximum−0.540.88
x-intercept00
Minimum1.87−6.06
x-intercept30

Step-by-step solution

6 steps

To sketch the graph of y=x3−2x2−3xy = x^3 - 2x^2 - 3x, we need to find the intercepts, turning points, and the point of inflection.

Continuing from 9.1: Determine the intercepts with the axes.

Step 2 (continued): Factor the quadratic x2−2x−3=0x^2 - 2x - 3 = 0.
We need two numbers that multiply to -3 and add to -2. These are -3 and 1.

x(x−3)(x+1)=0x(x-3)(x+1) = 0

This gives us the x-intercepts:

x=0orx−3=0orx+1=0x = 0 \quad \text{or} \quad x - 3 = 0 \quad \text{or} \quad x + 1 = 0 x=0orx=3orx=−1x = 0 \quad \text{or} \quad x = 3 \quad \text{or} \quad x = -1

The x-intercepts are (-1, 0), (0, 0), and (3, 0)\boxed{\text{(-1, 0), (0, 0), and (3, 0)}}

Part 9.2· 3 steps

Determine the coordinates of the turning points.

  1. Find the first derivative of the function.

    y=x3−2x2−3xy = x^3 - 2x^2 - 3x dydx=3x2−4x−3\frac{dy}{dx} = 3x^2 - 4x - 3

  2. Set the first derivative to zero to find the x-coordinates of the turning points.

    3x2−4x−3=03x^2 - 4x - 3 = 0

    Use the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}:

    x=−(−4)±(−4)2−4(3)(−3)2(3)x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-3)}}{2(3)} x=4±16+366x = \frac{4 \pm \sqrt{16 + 36}}{6} x=4±526x = \frac{4 \pm \sqrt{52}}{6} x=4±2136x = \frac{4 \pm 2\sqrt{13}}{6} x=2±133x = \frac{2 \pm \sqrt{13}}{3}

    This gives two x-coordinates for the turning points:

    x1=2+133≈2+3.6053≈5.6053≈1.868x_1 = \frac{2 + \sqrt{13}}{3} \approx \frac{2 + 3.605}{3} \approx \frac{5.605}{3} \approx 1.868 x2=2−133≈2−3.6053≈−1.6053≈−0.535x_2 = \frac{2 - \sqrt{13}}{3} \approx \frac{2 - 3.605}{3} \approx \frac{-1.605}{3} \approx -0.535

  3. Substitute these x-values back into the original function to find the corresponding y-coordinates.

    For x1≈1.868x_1 \approx 1.868:

    y1=(1.868)3−2(1.868)2−3(1.868)y_1 = (1.868)^3 - 2(1.868)^2 - 3(1.868) y1≈6.505−2(3.490)−5.604y_1 \approx 6.505 - 2(3.490) - 5.604 y1≈6.505−6.980−5.604y_1 \approx 6.505 - 6.980 - 5.604 y1≈−6.079y_1 \approx -6.079

    Turning Point 1: (1.87, -6.08)\boxed{\text{(1.87, -6.08)}} (local minimum)

    For x2≈−0.535x_2 \approx -0.535:

    y2=(−0.535)3−2(−0.535)2−3(−0.535)y_2 = (-0.535)^3 - 2(-0.535)^2 - 3(-0.535) y2≈−0.153−2(0.286)+1.605y_2 \approx -0.153 - 2(0.286) + 1.605 y2≈−0.153−0.572+1.605y_2 \approx -0.153 - 0.572 + 1.605 y2≈0.880y_2 \approx 0.880

    Turning Point 2: (-0.54, 0.88)\boxed{\text{(-0.54, 0.88)}} (local maximum)

Part 9.3· 3 steps

Determine the coordinates of the point of inflection.

  1. Find the second derivative of the function.

    dydx=3x2−4x−3\frac{dy}{dx} = 3x^2 - 4x - 3 d2ydx2=6x−4\frac{d^2y}{dx^2} = 6x - 4

  2. Set the second derivative to zero to find the x-coordinate of the point of inflection.

    6x−4=06x - 4 = 0 6x=46x = 4 x=46=23x = \frac{4}{6} = \frac{2}{3}

  3. Substitute this x-value back into the original function to find the corresponding y-coordinate.

    y=(23)3−2(23)2−3(23)y = \left(\frac{2}{3}\right)^3 - 2\left(\frac{2}{3}\right)^2 - 3\left(\frac{2}{3}\right) y=827−2(49)−2y = \frac{8}{27} - 2\left(\frac{4}{9}\right) - 2 y=827−89−2y = \frac{8}{27} - \frac{8}{9} - 2

    To combine these, find a common denominator (27):

    y=827−8×39×3−2×271×27y = \frac{8}{27} - \frac{8 \times 3}{9 \times 3} - \frac{2 \times 27}{1 \times 27} y=827−2427−5427y = \frac{8}{27} - \frac{24}{27} - \frac{54}{27} y=8−24−5427y = \frac{8 - 24 - 54}{27} y=−7027y = \frac{-70}{27} y≈−2.59y \approx -2.59

    The point of inflection is \boxed{\text{(\frac{2}{3}, -\frac{70}{27}) \approx (0.67, -2.59)}}

Part 9.4

Sketch the graph.

To sketch the graph, plot the following points:

  • x-intercepts: (−1,0)(-1, 0), (0,0)(0, 0), (3,0)(3, 0)
  • y-intercept: (0,0)(0, 0)
  • Local maximum: (−0.54,0.88)(-0.54, 0.88)
  • Local minimum: (1.87,−6.08)(1.87, -6.08)
  • Point of inflection: (0.67,−2.59)(0.67, -2.59)

Since it's a cubic function with a positive leading coefficient (x3x^3), the graph will:

  • Start from negative infinity on the left (as x→−∞,y→−∞x \to -\infty, y \to -\infty).
  • Rise to the local maximum at (−0.54,0.88)(-0.54, 0.88).
  • Decrease, passing through the y-intercept (0,0)(0,0) and the point of inflection (0.67,−2.59)(0.67, -2.59), to the local minimum at (1.87,−6.08)(1.87, -6.08).
  • Increase from the local minimum, passing through the x-intercept (3,0)(3,0), towards positive infinity on the right (as x→∞,y→∞x \to \infty, y \to \infty).

A sketch would look like this:
(Imagine an x-y coordinate plane)

  1. Mark the x-intercepts at -1, 0, and 3.
  2. Mark the y-intercept at 0.
  3. Plot the local maximum slightly to the left of the y-axis and above it.
  4. Plot the local minimum to the right of the y-axis and significantly below the x-axis.
  5. Draw a smooth curve connecting these points, starting from the bottom left, going up through (−1,0)(-1,0) to the local max, then down through (0,0)(0,0) and the point of inflection to the local min, and finally up through (3,0)(3,0) to the top right.

[Since I cannot draw an image, I have provided a detailed description of how to sketch the graph based on the calculated points.]

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