How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. #y=2x^2 + 1#?

x = -b/2a = 0 -> f(0) = 1. Coordinate of vertex: x = 0 and y = 1. Axis of symmetry: x = -b/2a = 0, then Oy is the axis of symmetry. x-intercepts: (x^2 + 1) is always positive, then there are no x-intercepts. The upward parabola doesn't intercept the x axis.

To graph the function ( y = 2x^2 + 1 ), follow these steps:

1. Plot the vertex: The vertex of the parabola is at the point ((h, k)), where (h = -\frac{b}{2a}) and (k) is the value of the function at (h).

2. Label the vertex: Once you've found the coordinates of the vertex, plot it on the graph.

3. Determine the axis of symmetry: The axis of symmetry is a vertical line passing through the vertex. Its equation is (x = h), where (h) is the x-coordinate of the vertex.

4. Plot the x-intercepts: To find the x-intercepts, set (y = 0) and solve for (x). These points will be where the parabola intersects the x-axis.

5. Label the x-intercepts: Once you've found the x-intercepts, plot them on the graph.

6. Draw the graph: Connect the vertex and the x-intercepts to form the parabola.

Given the function ( y = 2x^2 + 1 ):

1. Calculate the vertex: (a = 2), (b = 0), (c = 1) (h = -\frac{0}{2(2)} = 0) (k = 2(0)^2 + 1 = 1) So, the vertex is at ((0, 1)).

2. Label the vertex: Plot the point ((0, 1)) on the graph.

3. Determine the axis of symmetry: Since (h = 0), the axis of symmetry is the vertical line (x = 0).

4. Plot the x-intercepts: Set (y = 0): (0 = 2x^2 + 1) (2x^2 = -1) There are no real solutions, so there are no x-intercepts.

5. Draw the graph: With the vertex and axis of symmetry plotted, draw the parabola opening upwards.

Your graph should have a vertex at ((0, 1)), an axis of symmetry along (x = 0), and no x-intercepts.