To solve this problem, we need to analyze the quadratic equation given:

$y = x^2 + 2x - 3$

Step 1: Finding the Vertex

The vertex form of a quadratic function $y = ax^2 + bx + c$ can help us determine the vertex's coordinates.

For the equation $y = x^2 + 2x - 3$:

• $a = 1$
• $b = 2$
• $c = -3$

The x-coordinate of the vertex can be found using the formula: $x = -\frac{b}{2a}$ Substituting the values: $x = -\frac{2}{2 \cdot 1} = -1$

Now, substitute $x = -1$ back into the equation to find the y-coordinate: $y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$

So, the coordinates of the vertex are:

• $x = -1$
• $y = -4$

Step 2: Finding the x-Intercepts

To find the x-intercepts, we set $y = 0$ and solve for $x$ in the equation: $x^2 + 2x - 3 = 0$ This factors as: $(x + 3)(x - 1) = 0$ Setting each factor equal to zero gives: $x + 3 = 0 \Rightarrow x = -3$ $x - 1 = 0 \Rightarrow x = 1$

Summary of the Solution

• The x-coordinate of the vertex is $x = -1$.
• The y-coordinate of the vertex is $y = -4$.
• The x-intercepts are $x = -3$ and $x = 1$.

Answer:

The correct choice from the options provided is f. $x = -3$, $x = 1$.

Would you like further details on any part of this solution?

Here are five related questions to deepen your understanding:

1. How would the vertex change if the equation were $y = x^2 + 4x - 3$?
2. What is the impact on the vertex if we change the value of $a$ in $y = ax^2 + 2x - 3$?
3. How can you determine if a parabola opens upwards or downwards based on its equation?
4. Why is the vertex a maximum point in downward-opening parabolas?
5. How can you determine the y-intercept of a quadratic function?

Tip: For any quadratic equation $y = ax^2 + bx + c$, the y-intercept is simply the constant term $c$.