y=x2+2x-3

To analyze the quadratic function \( y = x^2 + 2x - 3 \), we can perform several tasks:

1. Identify the Coefficients:

• The quadratic is of the form \( y = ax^2 + bx + c \).
• Here, \( a = 1 \), \( b = 2 \), and \( c = -3 \).

2. Find the Vertex:

The vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} = -\frac{2}{2(1)} = -1 \] Now, substitute \( x = -1 \) into the original equation to find the y-coordinate: \[ y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4 \] So, the vertex is at \( (-1, -4) \).

3. Find the Axis of Symmetry:

The axis of symmetry for the quadratic function is the vertical line that passes through the vertex: \[ x = -1 \]

4. Find the y-intercept:

The y-intercept occurs when \( x = 0 \): \[ y = 0^2 + 2(0) - 3 = -3 \] So, the y-intercept is at \( (0, -3) \).

5. Find the x-intercepts (if any):

To find the x-intercepts, set \( y = 0 \): \[ 0 = x^2 + 2x - 3 \] Factoring the quadratic gives: \[ (x + 3)(x - 1) = 0 \] Thus, the x-intercepts are: \[ x = -3 \quad \text{and} \quad x = 1 \] So, the x-intercepts are at \( (-3, 0) \) and \( (1, 0) \).

6. Determine the Direction of Opening:

Since the coefficient \( a = 1 \) is positive, the parabola opens upwards.

7. Sketching the Graph:

From the information above:

• The vertex is at \( (-1, -4) \).
• The axis of symmetry is \( x = -1 \).
• The y-intercept is \( (0, -3) \).
• The x-intercepts are \( (-3, 0) \) and \( (1, 0) \).

You can plot these points and sketch a parabola that opens upwards.

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