You are asked to graph the function $f(x) = 2x^2 - 3$ and plot five points on the graph, including one point with $x = 0$, two with negative $x$-values, and two with positive $x$-values.

To graph this function, let's calculate some points by substituting different values of $x$ into the function $f(x) = 2x^2 - 3$:

1. $x = 0$:

$f(0) = 2(0)^2 - 3 = -3$ Point: $(0, -3)$

2. $x = -2$ (negative):

$f(-2) = 2(-2)^2 - 3 = 2(4) - 3 = 8 - 3 = 5$ Point: $(-2, 5)$

3. $x = -1$ (negative):

$f(-1) = 2(-1)^2 - 3 = 2(1) - 3 = 2 - 3 = -1$ Point: $(-1, -1)$

4. $x = 1$ (positive):

$f(1) = 2(1)^2 - 3 = 2(1) - 3 = 2 - 3 = -1$ Point: $(1, -1)$

5. $x = 2$ (positive):

$f(2) = 2(2)^2 - 3 = 2(4) - 3 = 8 - 3 = 5$ Point: $(2, 5)$

The five points to plot are:

• $(0, -3)$
• $(-2, 5)$
• $(-1, -1)$
• $(1, -1)$
• $(2, 5)$

You can now use these points to plot the graph on your grid. It will form a parabola opening upwards since the coefficient of $x^2$ is positive.

Would you like more details or have any questions?

5 Related Questions:

1. How do you determine the vertex of a parabola?
2. What is the axis of symmetry for this parabola?
3. How can you find the intercepts of the graph with the x-axis?
4. How would the graph change if the function were $f(x) = -2x^2 + 3$?
5. What are the steps to find the domain and range of this quadratic function?

Tip:

When graphing parabolas, always calculate points symmetrically around the vertex to ensure accuracy.