Question

Which of the following graphs represents the solution set of the quadratic equation?

y≤2x2+2x+1

Answer 1

To determine which graph represents the solution set of the quadratic inequality \( y \leq 2x^2 + 2x + 1 \), we need to analyze the quadratic \( 2x^2 + 2x + 1 \).

1. **Identify the Quadratic Function**:
The quadratic function is given by \( f(x) = 2x^2 + 2x + 1 \).

2. **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 \( \left(-\frac{1}{2}, \frac{1}{2}\right) \).

3. **Direction of the Parabola**:
Since the coefficient of \( x^2 \) (which is 2) is positive, the parabola opens upwards.

4. **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.

5. **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 \( (0, 1) \).

6. **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 \( y \leq 2x^2 + 2x + 1 \).