1. \( f(x)=x^{2}+4 x-3 \)

The function you provided is a quadratic function given by: \[ f(x) = x^2 + 4x - 3 \] ### Key Features of the Function 1. **Standard Form**: The function is already in standard form \( f(x) = ax^2 + bx + c \) where \( a = 1 \), \( b = 4 \), and \( c = -3 \). 2. **Vertex**: The vertex of a quadratic function in standard form can be found using the formula: \[ x = -\frac{b}{2a} \] Substituting the values of \( a \) and \( b \): \[ x = -\frac{4}{2 \cdot 1} = -2 \] To find the y-coordinate of the vertex, substitute \( x = -2 \) back into the function: \[ f(-2) = (-2)^2 + 4(-2) - 3 = 4 - 8 - 3 = -7 \] Thus, the vertex is at the point \( (-2, -7) \). 3. **Axis of Symmetry**: The axis of symmetry is the vertical line that passes through the vertex: \[ x = -2 \] 4. **Y-Intercept**: The y-intercept occurs when \( x = 0 \): \[ f(0) = 0^2 + 4(0) - 3 = -3 \] So, the y-intercept is at the point \( (0, -3) \). 5. **X-Intercepts**: To find the x-intercepts, set \( f(x) = 0 \): \[ x^2 + 4x - 3 = 0 \] We can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting \( a = 1 \), \( b = 4 \), and \( c = -3 \): \[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 12}}{2} = \frac{-4 \pm \sqrt{28}}{2} = \frac{-4 \pm 2\sqrt{7}}{2} = -2 \pm \sqrt{7} \] Thus, the x-intercepts are: \[ x = -2 + \sqrt{7} \quad \text{and} \quad x = -2 - \sqrt{7} \] ### Summary of Key Features - Vertex: \( (-2, -7) \) - Axis of Symmetry: \( x = -2 \) - Y-Intercept: \( (0, -3) \) - X-Intercepts: \( (-2 + \sqrt{7}, 0) \) and \( (-2 - \sqrt{7}, 0) \) If you have any specific questions or need further analysis on this function, feel free to ask!