1. f(x)=x^2+6x+8 Zero(s): __ Y-intercept: __ Axis of Symmetry: __ Vertex: __ Max or Min Domain: __ Range: __

1. The zeros of a function are the x-values where the function equals zero. To find the zeros of the function \(f(x) = x^2 + 6x + 8\), we set the function equal to zero and solve for x. This gives us the equation . Factoring this equation gives us \((x + 2)(x + 4) = 0\), which has solutions x = -2 and x = -4.2. The y-intercept of a function is the y-value where the function crosses the y-axis. This occurs when x = 0. Substituting x = 0 into the function gives us \(f(0) = 0^2 + 6(0) + 8 = 8\). Therefore, the y-intercept is 8.3. The axis of symmetry of a parabola is the vertical line that divides the parabola into two mirror images. For a quadratic function in the form \(f(x) = ax^2 + bx + c\), the axis of symmetry is given by the formula . In this case, a = 1 and b = 6, so the axis of symmetry is \(x = -6/2(1) = -3\).4. The vertex of a parabola is the point where the parabola reaches its maximum or minimum value. For a parabola that opens upwards, the vertex is the minimum point. The x-coordinate of the vertex is the same as the axis of symmetry, which is x = -3. Substituting x = -3 into the function gives us \(f(-3) = (-3)^2 + 6(-3) + 8 = -1\). Therefore, the vertex is (-3, -1), and since the parabola opens upwards, the vertex is a minimum.5. The domain of a function is the set of all possible x-values. For a quadratic function, the domain is all real numbers, or (-∞, ∞).6. The range of a function is the set of all possible y-values. Since the parabola opens upwards and has a minimum value at y = -1, the range is all y-values greater than or equal to -1, or [-1, ∞).