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For the curve y = -2x3 + 3x2, find the x-intercepts, y-intercept, and stationary points.

Mathematics

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3 parts
a(i)

(0,0) and (32,0)(0, 0) \text{ and } \left(\frac{3}{2}, 0\right)

a(ii)
(0, 0)
b

(0,0) and (1,1)(0, 0) \text{ and } (1, 1)

Graph of y = −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 y = −2x³ + 3x²
Pointxy
Minimum00
Maximum11
x-intercept1.50

Step-by-step solution

12 steps

here's the solution to question 23.

The equation of the curve is y=−2x3+3x2y = -2x^3 + 3x^2.

Part a(i)· 3 steps

the x-intercept of the curve.

Find

  1. Set y=0y=0 to find the x-intercepts.

    0=−2x3+3x20 = -2x^3 + 3x^2

  2. Factor out x2x^2.

    0=x2(−2x+3)0 = x^2(-2x + 3)

  3. Solve for xx.

    x2=0  ⟹  x=0x^2 = 0 \implies x = 0 −2x+3=0  ⟹  2x=3  ⟹  x=32-2x + 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}

    The x-intercepts are (0,0) and (32,0)\boxed{(0, 0) \text{ and } \left(\frac{3}{2}, 0\right)}.

Part a(ii)(0, 0)

the y-intercept of the curve.

  1. Set x=0x=0 to find the y-intercept.

    y=−2(0)3+3(0)2y = -2(0)^3 + 3(0)^2

  2. Calculate yy.

    y=0y = 0

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

Part b

(0,0) and (1,1)(0, 0) \text{ and } (1, 1)

(i) Determine the stationary points of the curve.

  1. Find the first derivative dydx\frac{dy}{dx}.

    y=−2x3+3x2y = -2x^3 + 3x^2 dydx=−6x2+6x\frac{dy}{dx} = -6x^2 + 6x

  2. Set dydx=0\frac{dy}{dx} = 0 to find the x-coordinates of the stationary points.

    −6x2+6x=0-6x^2 + 6x = 0

  3. Factor out −6x-6x.

    −6x(x−1)=0-6x(x - 1) = 0

  4. Solve for xx.

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

  5. Substitute the x-values back into the original equation y=−2x3+3x2y = -2x^3 + 3x^2 to find the corresponding y-values.

    For x=0x=0:

    y=−2(0)3+3(0)2=0y = -2(0)^3 + 3(0)^2 = 0

    Point: (0,0)(0, 0)

    For x=1x=1:

    y=−2(1)3+3(1)2=−2+3=1y = -2(1)^3 + 3(1)^2 = -2 + 3 = 1

    Point: (1,1)(1, 1)
    The stationary points are (0,0) and (1,1)\boxed{(0, 0) \text{ and } (1, 1)}.

Part b(ii)· 2 steps

For each point in (b)(i) above, determine whether it is a maximum or a minimum.

  1. Find the second derivative d2ydx2\frac{d^2y}{dx^2}.

    dydx=−6x2+6x\frac{dy}{dx} = -6x^2 + 6x d2ydx2=−12x+6\frac{d^2y}{dx^2} = -12x + 6

  2. Evaluate d2ydx2\frac{d^2y}{dx^2} at each stationary point.

    For (0,0)(0, 0) (where x=0x=0):

    d2ydx2=−12(0)+6=6\frac{d^2y}{dx^2} = -12(0) + 6 = 6

    Since d2ydx2>0\frac{d^2y}{dx^2} > 0, the point (0,0)(0, 0) is a local minimum.

    For (1,1)(1, 1) (where x=1x=1):

    d2ydx2=−12(1)+6=−12+6=−6\frac{d^2y}{dx^2} = -12(1) + 6 = -12 + 6 = -6

    Since d2ydx2<0\frac{d^2y}{dx^2} < 0, the point (1,1)(1, 1) is a local maximum.

Part c

Sketch the curve.

The curve has the following characteristics:

  • x-intercepts: (0,0)(0, 0) and (32,0)(\frac{3}{2}, 0)
  • y-intercept: (0,0)(0, 0)
  • Local minimum: (0,0)(0, 0)
  • Local maximum: (1,1)(1, 1)

The sketch should show a curve that:

  1. Comes from positive y-values (as x→−∞x \to -\infty, y→∞y \to \infty).
  2. Decreases to a local minimum at (0,0)(0, 0).
  3. Increases to a local maximum at (1,1)(1, 1).
  4. Decreases again, passing through the x-axis at (32,0)(\frac{3}{2}, 0) and continuing towards negative y-values (as x→∞x \to \infty, y→−∞y \to -\infty).

Here is a description of the sketch:
Draw a coordinate plane. Plot the points (0,0)(0,0), (1,1)(1,1), and (32,0)(\frac{3}{2},0).
The curve starts from the top-left, goes down to touch the origin (0,0)(0,0) as a minimum point. From the origin, it turns and goes up to the point (1,1)(1,1) as a maximum point. From (1,1)(1,1), it turns and goes down, crossing the x-axis at (32,0)(\frac{3}{2},0) and continuing downwards.

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