Y=x^2+4x-5
The equation you provided is a quadratic function in standard form, which can be rewritten as:
\[ y = ax^2 + bx + c \]
where:
- \( a = 1 \) (the coefficient of \( x^2 \)),
- \( b = 4 \) (the coefficient of \( x \)),
- \( c = -5 \) (the constant term).
Key Features of the Quadratic Function
Vertex: The vertex of a quadratic function in the form \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate:
\[ x = -\frac{b}{2a} \]
Substituting \( a \) and \( b \):
\[ x = -\frac{4}{2(1)} = -2 \]
To find the y-coordinate, substitute \( x = -2 \) back into the equation:
\[ y = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9 \]
Therefore, the vertex is at the point \( (-2, -9) \).
Axis of Symmetry: The axis of symmetry can be expressed as:
\[ x = -2 \]
Y-intercept: To find the y-intercept, set \( x = 0 \):
\[ y = (0)^2 + 4(0) - 5 = -5 \]
So the y-intercept is at the point \( (0, -5) \).
X-intercepts: To find the x-intercepts, set \( y = 0 \) and solve the equation:
\[ 0 = x^2 + 4x - 5 \]
This can be factored or solved using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-5)}}{2(1)} \] \[ x = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm \sqrt{36}}{2} = \frac{-4 \pm 6}{2} \]
This gives us the solutions:
\[ x = 1 \quad \text{and} \quad x = -5 \]
Therefore, the x-intercepts are at the points \( (1, 0) \) and \( (-5, 0) \).
Summary of Key Features
- Vertex: \( (-2, -9) \)
- Axis of Symmetry: \( x = -2 \)
- Y-intercept: \( (0, -5) \)
- X-intercepts: \( (1, 0) \) and \( (-5, 0) \)
If you need further details or specific information regarding this quadratic function, let me know!
