To graph the function \( y = 4x^3 - 6x^2 \), we start by analyzing and sketching its key components, including its form, intercepts, critical points, and overall behavior.
Step 1: Identify the function characteristics
- Form: This is a cubic polynomial. Its general shape will be determined by its leading term, \( 4x^3 \).
- Intercepts:
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Y-Intercept: Set \( x = 0 \): \[ y = 4(0)^3 - 6(0)^2 = 0 \] So the y-intercept is \( (0, 0) \).
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X-Intercepts: Set \( y = 0 \): \[ 4x^3 - 6x^2 = 0 \] Factor out \( 2x^2 \): \[ 2x^2(2x - 3) = 0 \] This gives the solutions: \[ 2x^2 = 0 \quad \Rightarrow \quad x = 0 \quad \text{(double root)} \] \[ 2x - 3 = 0 \quad \Rightarrow \quad x = \frac{3}{2} \] So, the x-intercepts are \( (0, 0) \) and \( \left( \frac{3}{2}, 0 \right) \).
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Step 2: Find critical points
To find critical points, we take the derivative \( y' \) and set it to zero: \[ y' = 12x^2 - 12x \] Factoring gives: \[ y' = 12x(x - 1) = 0 \] Setting this to zero:
- \( x = 0 \)
- \( x = 1 \)
Step 3: Determine the nature of critical points
To determine whether these critical points are maxima, minima, or points of inflection, we can use the second derivative: \[ y'' = 24x - 12 \]
- At \( x = 0 \): \[ y''(0) = 24(0) - 12 = -12 \quad \text{(local maximum)} \]
- At \( x = 1 \): \[ y''(1) = 24(1) - 12 = 12 \quad \text{(local minimum)} \]
Step 4: Calculate values at critical points
Now we can find the corresponding \( y \) values:
- At \( x = 0 \): \[ y(0) = 0 \]
- At \( x = 1 \): \[ y(1) = 4(1)^3 - 6(1)^2 = 4 - 6 = -2 \]
Step 5: Analyze end behavior
Since it is a cubic polynomial, as \( x \to -\infty \), \( y \to -\infty \) and as \( x \to +\infty \), \( y \to +\infty \).
Step 6: Sketch the graph
Using the information gathered, we can now sketch the graph:
- X-Intercepts: \( (0, 0) \) and \( \left( \frac{3}{2}, 0 \right) \)
- Critical points: Local maximum at \( (0,0) \) and local minimum at \( (1, -2) \)
- End behavior: Down left, up right.
The graph will start from the lower left, rise through the origin, reach a local maximum at (0, 0), then drop down to a local minimum at (1, -2), and finally rise again through \( \left( \frac{3}{2}, 0 \right) \) heading up towards positive infinity on the right.
You can plot these points and features on graph paper or using graphing software for a clearer representation.
If you have specific questions or need further assistance with this function, feel free to ask!