Complete the square for the following quadratic equation to determine its solutions and the location of its extreme valuey = -x + 4x + 12Ax=2 + 277extreme value at (2.16)B.X = -2,6extreme value at (2,12)C.x = -2.6extreme value at (2.16)D. X = 2 + 277extreme value at (2,12)

Accepted Solution

Answer:Option C. x = -2,6  extreme value at (2.16)Step-by-step explanation:we have[tex]y=-x^2+4x+12[/tex]This is the equation of a vertical parabola open downThe vertex is a maximum (extreme value)Convert the equation into vertex form[tex]y=-x^2+4x+12[/tex]Complete the squareGroup terms that contain the same variable and move the constant term to the left side[tex]y-12=-x^2+4x[/tex]Factor -1[tex]y-12=-(x^2-4x)[/tex]Remember to balance the equation by adding the same constants to each side.[tex]y-12-4=-(x^2-4x+4)[/tex]Rewrite as perfect squares[tex]y-16=-(x-2)^2[/tex][tex]y=-(x-2)^2+16[/tex] -----> equation of the parabola in vertex formThe vertex is the point (2,16) ----> is a maximum (extreme value)Determine the solutions of the quadratic equationFor y=0[tex]0=-(x-2)^2+16[/tex] [tex](x-2)^2=16[/tex] square root both sides[tex](x-2)=(+/-)4[/tex] [tex]x=2(+/-)4[/tex] [tex]x=2(+)4=6[/tex] [tex]x=2(-)4=-2[/tex] thereforeThe solutions are x=-2 and x=6The extreme value is (2,16)