Question 1005900

given that the axis of symmetry of the graph of the function {{{y=x^2+kx+5=-3}}}. Find the value of {{{k}}}.

recall: The axis of symmetry of a parabola is the vertical line through the vertex. For a parabola in standard form, {{{y = ax^2 + bx + c}}}, the axis of symmetry has the equation {{{x= -b/2a}}}.

Note that {{{-b/2a}}} is also the x-coordinate of the vertex of the parabola. 

if {{{x^2+kx+5=-3}}}, where {{{a=1}}} and {{{b=k}}} we have

{{{x= -k/2*1}}}

{{{x= -k/2}}} 

{{{2x= -k}}} 

{{{k=-2x}}}....substitute in given equation and solve for {{{x}}}


{{{x^2-2x*x+5=-3}}}

{{{x^2-2x^2+5=-3}}}

{{{-x^2+5=-3}}}

{{{3+5=x^2}}}

{{{8=x^2}}}

{{{x=sqrt(8)}}} 

{{{x=2sqrt(2)}}}
or {{{x=-2sqrt(2)}}}

then, go back to {{{k=-2x}}}, substitute values {{{2sqrt(2)}}} and {{{-2sqrt(2)}}} for {{{x}}} and find {{{k}}}


{{{k=-2x}}}=>{{{k=-2*2sqrt(2)}}}=>{{{k=-4sqrt(2)}}}
or

{{{k=-2x}}}=>{{{k= -2*(-2sqrt(2))}}}=>{{{k=4sqrt(2)}}}

so, your equations are:

{{{y=x^2-4sqrt(2)x+5}}} or {{{y=x^2+4sqrt(2)x+5}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
line(2sqrt(2),-10,2sqrt(2),10),line(-2sqrt(2),-10,-2sqrt(2),10),
 graph( 600, 600, -10, 10, -10, 10, x^2-4sqrt(2)x+5, x^2+4sqrt(2)x+5)) }}}