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Given the function f(x) = -2x3 - 3x2, find the x-intercepts, y-intercept, and turning points.

Mathematics

ScanSolve AI Answer

Graph of f(x) = −2x³ − 3x²

Use the left and right arrow keys to trace the curve, up and down to jump between key points, and plus and minus to zoom.

Key points of f(x) = −2x³ − 3x²
Pointxy
x-intercept−1.50
Minimum−1−1
Maximum00

Step-by-step solution

3 steps

Here are the solutions to Question 7:

Given the function f(x)=−2x3−3x2f(x) = -2x^3 - 3x^2.

Part 7.1

To determine the x-intercepts, we set f(x)=0f(x) = 0.

−2x3−3x2=0-2x^3 - 3x^2 = 0

Factor out the common term x2x^2:

−x2(2x+3)=0-x^2(2x + 3) = 0

This gives two possible solutions:

−x2=0  ⟹  x=0-x^2 = 0 \implies x = 0 2x+3=0  ⟹  2x=−3  ⟹  x=−322x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2}

The x-intercepts are \boxed{\text{(0, 0) and (-\frac{3}{2}, 0)}}.

Part 7.2

To determine the y-intercept, we set x=0x = 0.

f(0)=−2(0)3−3(0)2f(0) = -2(0)^3 - 3(0)^2 f(0)=0−0f(0) = 0 - 0 f(0)=0f(0) = 0

The y-intercept is (0, 0)\boxed{\text{(0, 0)}}.

Part 7.3· 3 steps

To determine the coordinates of the turning points, we first find the first derivative f′(x)f'(x) and set it to zero.

  1. Find the derivative f′(x)f'(x).

    f′(x)=ddx(−2x3−3x2)f'(x) = \frac{d}{dx}(-2x^3 - 3x^2) f′(x)=−2(3)x3−1−3(2)x2−1f'(x) = -2(3)x^{3-1} - 3(2)x^{2-1} f′(x)=−6x2−6xf'(x) = -6x^2 - 6x

  2. Set f′(x)=0f'(x) = 0 to find the x-coordinates of the turning points.

    −6x2−6x=0-6x^2 - 6x = 0

    Factor out −6x-6x:

    −6x(x+1)=0-6x(x + 1) = 0

    This gives two possible solutions for xx:

    −6x=0  ⟹  x=0-6x = 0 \implies x = 0 x+1=0  ⟹  x=−1x + 1 = 0 \implies x = -1

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

    For x=0x = 0:

    f(0)=−2(0)3−3(0)2=0f(0) = -2(0)^3 - 3(0)^2 = 0

    Turning point 1: (0,0)(0, 0)

    For x=−1x = -1:

    f(−1)=−2(−1)3−3(−1)2f(-1) = -2(-1)^3 - 3(-1)^2 f(−1)=−2(−1)−3(1)f(-1) = -2(-1) - 3(1) f(−1)=2−3f(-1) = 2 - 3 f(−1)=−1f(-1) = -1

    Turning point 2: (−1,−1)(-1, -1)

    The coordinates of the turning points are (0, 0) and (-1, -1)\boxed{\text{(0, 0) and (-1, -1)}}.

Part 7.4

To sketch the graph of f(x)=−2x3−3x2f(x) = -2x^3 - 3x^2, we use the information gathered:

  • x-intercepts: (0,0)(0, 0) and (−32,0)(-\frac{3}{2}, 0) (or (−1.5,0)(-1.5, 0))
  • y-intercept: (0,0)(0, 0)
  • Turning points: (0,0)(0, 0) (local maximum) and (−1,−1)(-1, -1) (local minimum)

The function is a cubic with a negative leading coefficient, so it generally starts from the top-left, goes down, turns up, then turns down to the bottom-right.

The graph will:

  1. Start from the top-left.
  2. Pass through the x-intercept (−32,0)(-\frac{3}{2}, 0).
  3. Continue downwards to reach the local minimum at (−1,−1)(-1, -1).
  4. Turn upwards from the local minimum.
  5. Reach the local maximum at (0,0)(0, 0), which is also an x-intercept and the y-intercept.
  6. Turn downwards from the local maximum and continue to the bottom-right.
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