A quadratic of the form $-2x^2 + bx + c$ has roots of x=3+sqrt5 and x=3-sqrt5 The graph of
$y = -2x^2 + bx + c$
is a parabola. Find the vertex of this parabola.
Hint-The coefficient of x^2 is not 1!
+23246 If the roots are 3 + sqrt(5) and 3 - sqrt(5), then the equation is:
y = [x - ( 3 + sqrt(5) ) ] · [x - ( 3 - sqrt(5) ) ]
If you expand the above equation, you will end with: y = x2 - 6x + 4
Now, to find the vertex: y = x2 - 6x + 4
y - 4 = x2 - 6x
Complete the square: y - 4 + 9 = x2 - 6x + 9
y + 5 = (x - 3)2
So, the vertex is: (3, -5)
