\( f(x)=x^{2}+4 x-1 \) (b) Find the vertex and \( x \) - and \( y \)-intercepts of \( f \). (If an answer does not exist, enter DN vertex \( (x, y)=(\square) \) x-intercepts \( \quad(x, y)=( \) smaller \( x \)-value)

To find the vertex and the intercepts of the function \( f(x) = x^2 + 4x - 1 \), we can follow these steps: ### 1. Finding the Vertex The vertex of a quadratic function in the form \( f(x) = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \] In this case, \( a = 1 \) and \( b = 4 \): \[ x = -\frac{4}{2 \cdot 1} = -2 \] Now, we can find the y-coordinate of the vertex by substituting \( x = -2 \) back into the function: \[ f(-2) = (-2)^2 + 4(-2) - 1 = 4 - 8 - 1 = -5 \] Thus, the vertex is: \[ \text{Vertex } (x, y) = (-2, -5) \] ### 2. Finding the x-intercepts To find the x-intercepts, we set \( f(x) = 0 \): \[ x^2 + 4x - 1 = 0 \] We can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting \( a = 1 \), \( b = 4 \), and \( c = -1 \): \[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm \sqrt{20}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5} \] This gives us two x-intercepts: \[ x_1 = -2 + \sqrt{5}, \quad x_2 = -2 - \sqrt{5} \] To find the smaller x-value: \[ x_2 = -2 - \sqrt{5} \] ### 3. Finding the y-intercept To find the y-intercept, we set \( x = 0 \): \[ f(0) = 0^2 + 4(0) - 1 = -1 \] Thus, the y-intercept is: \[ (0, -1) \] ### Summary of Results - Vertex: \( (x, y) = (-2, -5) \) - x-intercepts: \( (x, y) = (-2 - \sqrt{5}, 0) \) (smaller x-value) - y-intercept: \( (0, -1) \) So, the final answers are: - Vertex: \( (x, y) = (-2, -5) \) - x-intercepts: \( (x, y) = (-2 - \sqrt{5}, 0) \)