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).

a, b, c, d are correct.

step-by-step explanation:

we have g(x) = 3 and are required to find which option is a translation of g(x).

now, 'translation' is a transformation which means to shift the function in any direction and the general form of translation is g(x) + k, where k is any real number.

so, according to the options we see that all are correct as they satisfy the general form of translation.

also, it can be seen from the graph below that the function g(x) = 3 is shifted from its position in all the four options.