To analyze the end behavior of the function \( j(x) = \frac{12x^2}{5} - \frac{27x}{5} \), we start by rewriting it in standard polynomial form:
\[ j(x) = \frac{12}{5} x^2 - \frac{27}{5} x \]
This is a quadratic polynomial, specifically a parabola of the form \( ax^2 + bx + c \) where:
- \( a = \frac{12}{5} \)
- \( b = -\frac{27}{5} \)
- \( c = 0 \)
End Behavior:
The end behavior of a polynomial function is determined primarily by the leading term (the term with the highest degree). Since the leading term in this case is \( \frac{12}{5} x^2 \) (which has a positive coefficient), we can analyze the behavior as \( x \) approaches positive and negative infinity.
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As \( x \to +\infty \):
- \( j(x) \to +\infty \) because the quadratic term dominates and is positive.
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As \( x \to -\infty \):
- \( j(x) \to +\infty \) for the same reason, since a squared value (like \( x^2 \)) will still be positive.
Conclusion on End Behavior:
Thus, the end behavior of the graph of \( j(x) \) is:
- As \( x \to +\infty \), \( j(x) \to +\infty \)
- As \( x \to -\infty \), \( j(x) \to +\infty \)
This indicates that the parabola opens upward.
Graph Sketching:
To visualize the function, follow these steps to complete the sketch:
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Finding the Zeros: To find the zeros of \( j(x) \), we set the equation equal to zero:
\[ \frac{12}{5} x^2 - \frac{27}{5} x = 0 \]
Factoring out \( \frac{1}{5} x \):
\[ \frac{1}{5} x(12x - 27) = 0 \]
This gives us the solutions:
- \( x = 0 \)
- \( 12x - 27 = 0 \) → \( x = \frac{27}{12} = \frac{9}{4} = 2.25 \)
So, the zeros of the function are \( x = 0 \) and \( x = 2.25 \).
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Sketch the Graph:
- Plot these zeros on the x-axis.
- Since the parabola opens upwards and both ends go to positive infinity, sketch the graph that starts in the second quadrant (as it approaches positive infinity for negative \( x \)), passes through the zeros at \( (0, 0) \) and \( (2.25, 0) \), dips below (the vertex of the parabola), and returns back up towards positive infinity in the first quadrant.
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Vertex: Since the function is a quadratic, it has a vertex which can be found at \( x = -\frac{b}{2a} \): \[ x = -\frac{-\frac{27}{5}}{2 \cdot \frac{12}{5}} = \frac{27}{24} = 1.125 \] Calculating \( j(1.125) \): \[ j(1.125) = \frac{12}{5}(1.125^2) - \frac{27}{5}(1.125) \] This gives the vertex point which can be calculated numerically.
By now, you should have a basic graph of a parabola starting high on the left, dipping down through the zeros, reaching a minimum, and rising again high on the right. This completes the sketch of the graph of \( j(x) \), illustrating both the zeros and the end behavior.
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