To analyze the function \( y = 2x^2 + 4x - 1 \), we can determine the following features:
A) Vertex: The vertex of a quadratic function in the form \( y = ax^2 + bx + c \) can be found using the formula: \[ x = -\frac{b}{2a} \] In this case, \( a = 2 \) and \( b = 4 \): \[ x = -\frac{4}{2 \times 2} = -\frac{4}{4} = -1 \] Now we substitute \( x = -1 \) back into the function to find the y-coordinate of the vertex: \[ y = 2(-1)^2 + 4(-1) - 1 = 2(1) - 4 - 1 = 2 - 4 - 1 = -3 \] So, the vertex is: \[ \text{Vertex: } (-1, -3) \]
B) Axis of Symmetry: The axis of symmetry for a parabola is the vertical line that passes through the vertex. This is the \( x \)-coordinate of the vertex: \[ \text{Axis of Symmetry: } x = -1 \]
C) Minimum: Since the coefficient of \( x^2 \) (which is \( a = 2 \)) is positive, the parabola opens upwards, and the vertex represents the minimum point of the function. The minimum value is the y-coordinate of the vertex: \[ \text{Minimum: } -3 \]
D) Y-Intercept: To find the y-intercept of the function, set \( x = 0 \): \[ y = 2(0)^2 + 4(0) - 1 = -1 \] Thus, the y-intercept is: \[ \text{Y-Intercept: } -1 \]
Now, summarizing all the answers:
A) Vertex: \((-1, -3)\)
B) Axis of Symmetry: \(x = -1\)
C) Minimum: \(-3\)
D) Y-Intercept: \(-1\)