To determine which graph represents the solution set of the quadratic inequality \( y \leq 2x^2 + 2x + 1 \), we need to analyze the quadratic \( 2x^2 + 2x + 1 \).
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Identify the Quadratic Function: The quadratic function is given by \( f(x) = 2x^2 + 2x + 1 \).
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Finding the Vertex: The vertex of a quadratic function in the form \( ax^2 + bx + c \) can be found using the formula: \[ x = -\frac{b}{2a} \] Here, \( a = 2 \) and \( b = 2 \): \[ x = -\frac{2}{2 \cdot 2} = -\frac{1}{2} \] Now, substitute \( x = -\frac{1}{2} \) back into the function to find the y-coordinate of the vertex: \[ f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) + 1 = 2\left(\frac{1}{4}\right) - 1 + 1 = \frac{1}{2} \] Therefore, the vertex of the parabola is at \( \left(-\frac{1}{2}, \frac{1}{2}\right) \).
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Direction of the Parabola: Since the coefficient of \( x^2 \) (which is 2) is positive, the parabola opens upwards.
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Determine the Solution Set: The inequality \( y \leq 2x^2 + 2x + 1 \) represents all points on or below the parabola. This means we need to shade the region below the curve.
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Y-intercept: To find the y-intercept, set \( x=0 \): \[ f(0) = 2(0)^2 + 2(0) + 1 = 1 \] Therefore, the y-intercept is at point \( (0, 1) \).
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X-intercepts: To find the x-intercepts, set \( y=0 \): \[ 0 = 2x^2 + 2x + 1 \] The discriminant of the quadratic \( 2x^2 + 2x + 1 \) is: \[ b^2 - 4ac = 2^2 - 4(2)(1) = 4 - 8 = -4 \] Since the discriminant is negative, the quadratic has no real roots, indicating that the parabola does not touch the x-axis.
Now that we have analyzed the quadratic function and its behavior, you can look for a graph that displays an upward-opening parabola with its vertex at \( (-\frac{1}{2}, \frac{1}{2}) \), does not intersect the x-axis, and has shading below the curve representing the area where \( y \leq 2x^2 + 2x + 1 \).