Ineed with these algebra 2 questions. (60 points) 1. solve for x x^2+6x+11=0 2. find the vertex and axis of symmetry f(x)=2x^2-16x-17 3. given the equation f(x)=-2(x-3)^2+4 -how do we determine the vertex? -what direction is the graph opening? -is the vertex a maximum or a minimum?

Question 1:  x = -3 + i√2 and x = -3 - i√2

Question 2:  Vertex at (-3,2) and axis of symmetry x = -3.

Question 3:  See explanation below.

Step-by-step explanation:

x^2+6x+11=0 is a quadratic equation with coefficients a=1, b=6 and c=11.  Use the quadratic formula to determine the roots:  First, find the determinant (b^2-4ac); here it is 36-4(1)(11) = 36 - 44, or -8.  Because the determinant is negative, the roots (solutions) are complex, not real:

      -6 plus or minus i√8        -6 plus or minus 2i√2      

x = =

                     2                                   2

These reduce to:  x = -3 + i√2 and x = -3 - i√2

Question 2:  The equation of the axis of symmetry is x = -b/(2a).  Here, we get x = -6/(2*1), or x = -3.  The value of this function at x = -3 is 2.  Thus, the vertex is at (-3,2).

Question 3:  Compare the given f(x) = -2(x-3)^2 + 4 to y = a(x-h)^2 + k.  Identify the value of h as being 3 and that of k as being 4.  Thus, the vertex is at (3, 4).  the graph opens DOWN because the coefficient -2 is negative.  There's a maximum at (3,4).

i believe the answer is c.