Graph the equation y

To graph the equation $y = x^2 - 3$ by plotting points, we can generate a table of values for $x$ and compute the corresponding $y$ values. We'll choose several values for $x$, compute $y$, and then plot those points.

Step-by-Step Process:

1. Choose values for $x$. Let's select some values of $x$, both negative and positive, to get a good range for the graph.
2. Compute corresponding $y$ values using the equation $y = x^2 - 3$.
3. Plot the points on a coordinate plane.

Table of values:

$x$	$y = x^2 - 3$
-3	$(-3)^2 - 3 = 9 - 3 = 6$
-2	$(-2)^2 - 3 = 4 - 3 = 1$
-1	$(-1)^2 - 3 = 1 - 3 = -2$
0	$0^2 - 3 = 0 - 3 = -3$
1	$1^2 - 3 = 1 - 3 = -2$
2	$2^2 - 3 = 4 - 3 = 1$
3	$3^2 - 3 = 9 - 3 = 6$

Plot the points:

• (-3, 6)
• (-2, 1)
• (-1, -2)
• (0, -3)
• (1, -2)
• (2, 1)
• (3, 6)

These points can be plotted on a graph. When you connect them, you'll notice that the graph forms a parabola opening upwards with its vertex at $(0, -3)$.

Would you like me to create the graph for you, or do you have any questions about the process?

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Follow-up Questions:

1. What does the vertex of a parabola represent in the context of this equation?
2. How would the graph change if the equation was $y = x^2 + 3$?
3. What is the general shape of the graph for any quadratic equation of the form $y = x^2 + c$?
4. Can you explain the concept of symmetry in a parabola?
5. How do you find the vertex of a quadratic function algebraically?

Tip:

For quick graphing of quadratics, remember that the vertex $(h, k)$ of a parabola given by $y = a(x - h)^2 + k$ is the point where the axis of symmetry intersects the graph. If the equation is in standard form $y = x^2 + c$, the vertex is at $(0, c)$.