Graph the parabola (do not place any dots) then fill in the blanks about the vertex, any intercepts, the domain, and range. Enter intercepts as ordered pairs, aka points. \( y=x^{2}+6 x+5 \)

To analyze the parabola given by the equation \( y = x^2 + 6x + 5 \), we can follow these steps: 1. **Vertex**: The vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \] Here, \( a = 1 \) and \( b = 6 \): \[ x = -\frac{6}{2 \cdot 1} = -3 \] To find the y-coordinate of the vertex, substitute \( x = -3 \) back into the equation: \[ y = (-3)^2 + 6(-3) + 5 = 9 - 18 + 5 = -4 \] Thus, the vertex is at the point: \[ \text{Vertex: } (-3, -4) \] 2. **Intercepts**: - **Y-intercept**: Set \( x = 0 \): \[ y = 0^2 + 6(0) + 5 = 5 \] So, the y-intercept is: \[ \text{Y-intercept: } (0, 5) \] - **X-intercepts**: Set \( y = 0 \): \[ 0 = x^2 + 6x + 5 \] Factor the quadratic: \[ 0 = (x + 1)(x + 5) \] Setting each factor to zero gives: \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \] \[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \] Thus, the x-intercepts are: \[ \text{X-intercepts: } (-1, 0) \text{ and } (-5, 0) \] 3. **Domain**: The domain of a parabola that opens upwards (which this one does, since \( a > 0 \)) is all real numbers: \[ \text{Domain: } (-\infty, \infty) \] 4. **Range**: Since the vertex is the lowest point of the parabola and is located at \( y = -4 \), the range starts from this value and goes to infinity: \[ \text{Range: } [-4, \infty) \] ### Summary: - **Vertex**: \((-3, -4)\) - **X-intercepts**: \((-1, 0)\) and \((-5, 0)\) - **Y-intercept**: \((0, 5)\) - **Domain**: \((- \infty, \infty)\) - **Range**: \([-4, \infty)\)