To graph the quadratic function \( y = 4x^2 - 3x + 2 \) and identify the y-value of the y-intercept, we can follow these steps:
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Identify the y-intercept: The y-intercept occurs when \( x = 0 \). We can find the y-value by substituting \( x = 0 \) into the equation:
\[ y = 4(0)^2 - 3(0) + 2 = 2 \]
So, the y-value of the y-intercept is \( 2 \).
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Graphing the quadratic:
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The quadratic function can be rewritten in standard form \( y = ax^2 + bx + c \), where \( a = 4 \), \( b = -3 \), and \( c = 2 \).
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The coefficient \( a > 0 \) indicates that the parabola opens upwards.
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To sketch the graph, we can find the vertex and additional points:
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Finding the vertex: The x-coordinate of the vertex can be found using the formula \( x = -\frac{b}{2a} \):
\[ x = -\frac{-3}{2(4)} = \frac{3}{8} \]
Now substitute back to find the y-coordinate of the vertex:
\[ y = 4\left(\frac{3}{8}\right)^2 - 3\left(\frac{3}{8}\right) + 2 \]
\[ y = 4\left(\frac{9}{64}\right) - \frac{9}{8} + 2 = \frac{36}{64} - \frac{72}{64} + \frac{128}{64} \]
\[ y = \frac{36 - 72 + 128}{64} = \frac{92}{64} = \frac{23}{16} \]
Thus, the vertex is at \( \left(\frac{3}{8}, \frac{23}{16}\right) \).
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Plotting points: We can substitute some values of \( x \) to find additional points on the graph:
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For \( x = 1 \): \[ y = 4(1)^2 - 3(1) + 2 = 4 - 3 + 2 = 3 \]
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For \( x = -1 \): \[ y = 4(-1)^2 - 3(-1) + 2 = 4 + 3 + 2 = 9 \]
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For \( x = 2 \): \[ y = 4(2)^2 - 3(2) + 2 = 16 - 6 + 2 = 12 \]
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Sketch the graph: Now you can plot the points, including the vertex and the y-intercept. Draw a smooth curve through these points to form a parabola. The y-intercept is on the y-axis at the point \( (0, 2) \).
In summary, the y-value of the y-intercept is \( 2 \).