Question

PLEASE HELP! I need help 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?

Answer 1

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.

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).