Y=x^2+4x-5

The expression \( y = x^2 + 4x - 5 \) represents a quadratic function. Here’s a breakdown of some key features of this quadratic equation:

1. Standard Form:

The equation is already in standard form \( y = ax^2 + bx + c \), where:

• \( a = 1 \)
• \( b = 4 \)
• \( c = -5 \)

2. Vertex:

The vertex of a parabola given in standard form can be found using the vertex formula: \[ x = -\frac{b}{2a} \] Substituting the values of \( b \) and \( a \): \[ x = -\frac{4}{2 \cdot 1} = -2 \] To find the \( y \)-coordinate of the vertex, substitute \( x = -2 \) back into the equation: \[ y = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9 \] Thus, the vertex is at \( (-2, -9) \).

3. Axis of Symmetry:

The axis of symmetry for the parabola can be expressed by the line: \[ x = -2 \]

4. Y-Intercept:

To find the y-intercept, substitute \( x = 0 \) into the equation: \[ y = 0^2 + 4(0) - 5 = -5 \] So, the y-intercept is \( (0, -5) \).

5. X-Intercepts:

To find the x-intercepts, set \( y = 0 \) and solve for \( x \): \[ 0 = x^2 + 4x - 5 \] This can be factored or solved using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting in the values: \[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm \sqrt{36}}{2} = \frac{-4 \pm 6}{2} \] This gives us: \[ x = \frac{2}{2} = 1 \quad \text{and} \quad x = \frac{-10}{2} = -5 \] Thus, the x-intercepts are \( (1, 0) \) and \( (-5, 0) \).

6. Graph:

The graph of this quadratic function will be a parabola that opens upwards (since \( a > 0 \)) with a vertex at \( (-2, -9) \) and intercepts at \( (0, -5) \), \( (1, 0) \), and \( (-5, 0) \).

If you have any specific questions or need further analysis, feel free to ask!