Vertex: (1, 4)
Axis of symmetry: x = 1
X-intercepts: approximately (2.41, 0) and (-0.41, 0)
Y-intercept: (0, 2)
Vertex, Axis of Symmetry, and Intercepts for y = -2x^2 + 4x + 2
Vertex:
The vertex of a parabola can be found using the formula x = -b / 2a, where a and b are the coefficients of the x^2 and x terms, respectively.
In this equation, a = -2 and b = 4.
Therefore, the vertex's x-coordinate is x = -4 / (2 * -2) = 1.
To find the vertex's y-coordinate, substitute the x-coordinate (1) back into the equation: y = -2(1)^2 + 4(1) + 2 y = -2 + 4 + 2 y = 4
Therefore, the vertex is at (1, 4).
Axis of Symmetry:
The axis of symmetry of a parabola is a vertical line that passes through the vertex.
Since the vertex is at (1, 4), the axis of symmetry is the line x = 1.
X-Intercepts:
The x-intercepts occur where the parabola crosses the x-axis. This happens when y = 0.
Set the equation equal to 0 and solve for x: 0 = -2x^2 + 4x + 2 Use the quadratic formula to solve: x = (-b ± √(b^2 - 4ac)) / 2a x = (-4 ± √(4^2 - 4 * -2 * 2)) / 2 * -2 x = (-4 ± √(28)) / -4 x = (-4 ± 2√7) / -4 x = 1 ± √7 / 2
Therefore, the x-intercepts are approximately at (2.41, 0) and (-0.41, 0).
Y-Intercept:
The y-intercept occurs where the parabola crosses the y-axis. This happens when x = 0.
Set the equation equal to 0 and solve for x: y = -2(0)^2 + 4(0) + 2 y = 2
Therefore, the y-intercept is at (0, 2).