Let's analyze the function given in the image:

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

1. X- and Y-Intercepts

• Y-Intercept: Substitute $x = 0$ into the equation to find the y-intercept. $y = (0)^2 + 2(0) - 3 = -3$ So, the y-intercept is $(0, -3)$.

• X-Intercepts: Set $y = 0$ and solve for $x$. $x^2 + 2x - 3 = 0$ Factor this equation: $(x + 3)(x - 1) = 0$ Thus, $x = -3$ and $x = 1$. So, the x-intercepts are $(-3, 0)$ and $(1, 0)$.

2. Axis of Symmetry

The axis of symmetry for a quadratic function $y = ax^2 + bx + c$ is given by the formula: $x = -\frac{b}{2a}$ For our function, $a = 1$ and $b = 2$, so: $x = -\frac{2}{2 \times 1} = -1$ The axis of symmetry is $x = -1$.

3. Vertex

To find the vertex, substitute $x = -1$ into the equation to find the corresponding y-value: $y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$ So, the vertex is $(-1, -4)$.

4. Domain

Since this is a quadratic function (a parabola), the domain is all real numbers: $\text{Domain: } (-\infty, \infty)$

5. Range

Since the parabola opens upwards (the coefficient of $x^2$ is positive), the range starts at the y-coordinate of the vertex and goes to infinity: $\text{Range: } [-4, \infty)$

Would you like further clarification on any part?

Here are five related questions to deepen understanding:

1. How do you determine the vertex of a quadratic function if it's given in standard form?
2. What is the significance of the axis of symmetry in a parabola?
3. Can you find the domain and range of a quadratic function if it opens downwards?
4. How do you confirm x- and y-intercepts using a graphing calculator?
5. What are the general shapes of parabolas based on the sign of the leading coefficient?

Tip: When finding intercepts, always remember to substitute the opposite variable to zero. For example, for the y-intercept, let $x = 0$, and for x-intercepts, set $y = 0$.