To graph the quadratic function $y = x^2 + 2x - 3$, we can calculate five points by choosing different values of $x$ and calculating the corresponding values of $y$. Let's compute the points:

Step 1: Choose values of $x$

We'll use the following values of $x$: $-3, -2, -1, 0, 1$.

Step 2: Calculate the corresponding values of $y$

1. For $x = -3$: $y = (-3)^2 + 2(-3) - 3 = 9 - 6 - 3 = 0$ So, the point is $(-3, 0)$.

2. For $x = -2$: $y = (-2)^2 + 2(-2) - 3 = 4 - 4 - 3 = -3$ So, the point is $(-2, -3)$.

3. For $x = -1$: $y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$ So, the point is $(-1, -4)$.

4. For $x = 0$: $y = (0)^2 + 2(0) - 3 = 0 + 0 - 3 = -3$ So, the point is $(0, -3)$.

5. For $x = 1$: $y = (1)^2 + 2(1) - 3 = 1 + 2 - 3 = 0$ So, the point is $(1, 0)$.

Step 3: List the points

The five points are:

• $(-3, 0)$
• $(-2, -3)$
• $(-1, -4)$
• $(0, -3)$
• $(1, 0)$

Graphing these points:

These points represent a parabola opening upwards with a vertex at $(-1, -4)$.

Would you like to see a graph of this function, or do you have any questions? Let me know!

Related questions:

1. How do you find the vertex of a quadratic function?
2. What is the axis of symmetry of this quadratic function?
3. How can you determine if a parabola opens upwards or downwards?
4. How do you solve for the roots (x-intercepts) of the quadratic function?
5. What is the general form of a quadratic function, and how do you identify the coefficients?

Tip:

To easily find the vertex of a quadratic equation in the form $y = ax^2 + bx + c$, use the formula for the x-coordinate of the vertex:
$x = \frac{-b}{2a}$