Let's break down the problem.

Function:

You are given the quadratic function:

$f(x) = 2x^2 + 1$

1. Graphing the Function:

The function $f(x) = 2x^2 + 1$ is a parabola, and since the coefficient of $x^2$ is positive (2), the parabola opens upwards.

• The vertex of the parabola is at $(0, 1)$, as the constant term 1 represents the y-coordinate of the vertex.
• The parabola is vertically stretched because of the factor of 2 in front of $x^2$, making it narrower than the standard parabola $x^2$.
• For points near the vertex:
  • At $x = 0$, $f(0) = 2(0)^2 + 1 = 1$.
  • At $x = 1$, $f(1) = 2(1)^2 + 1 = 3$.
  • At $x = -1$, $f(-1) = 2(-1)^2 + 1 = 3$.
  • At $x = 2$, $f(2) = 2(2)^2 + 1 = 9$.
  • At $x = -2$, $f(-2) = 2(2)^2 + 1 = 9$.

You can plot these points: $(0, 1)$, $(1, 3)$, $(-1, 3)$, $(2, 9)$, and $(-2, 9)$ on the provided grid.

2. Domain:

The domain of the quadratic function $f(x) = 2x^2 + 1$ is all real numbers since the function is defined for all $x$.

$\text{Domain}: (-\infty, \infty)$

3. Range:

The minimum value of the function occurs at the vertex $(0, 1)$, and since the parabola opens upwards, the range is all values of $y \geq 1$.

$\text{Range}: [1, \infty)$

Would you like me to clarify any specific steps or plot the graph with further detail?

Here are 5 related questions:

1. How do you find the vertex of a quadratic function?
2. What impact does the coefficient in front of $x^2$ have on the shape of the parabola?
3. How would the graph change if the equation were $f(x) = -2x^2 + 1$?
4. What is the axis of symmetry for the function $f(x) = 2x^2 + 1$?
5. How would you determine the y-intercept and x-intercepts of this function?

Tip: For quadratic functions, the vertex form $f(x) = a(x-h)^2 + k$ can help identify the vertex and the direction in which the parabola opens easily.