Question 980730
{{{f(t) = t^2 + 6t -20}}}

Part A: Rewrite the function in vertex form by completing the square.

{{{f(t) = (t^2 + 6t+b^2)-b^2 -20}}}......recall that {{{a+b)^2=a^2+2ab+b^2}}}, and compare it to {{{f(t) = t^2 + 6t - 20}}}, you see that {{{a=1}}} and {{{2ab=6}}}; so, {{{2ab=6}}}=>{{{2*1b=6}}}=>{{{2b=6}}}=>{{{b=3}}}

than we have

{{{f(t) = (t^2 + 6t+3^2)-3^2 - 20}}}

{{{f(t) = (t + 3)^2-9 - 20}}}

{{{f(t) = (t + 3)^2 -29}}}

Part B: Determine the vertex and indicate whether it is a maximum or a minimum on the graph.

compare {{{f(t) = (t + 3)^2 -29}}} to vertex form of parabola {{{y=(x-h)^2+k}}}, you see that {{{h=-3}}} and {{{k=-29}}}  

so, the vertex is at:

({{{-3}}},{{{-29}}})


{{{drawing( 600, 600, -20, 20, -40, 40,
circle(-3,-29,.2),locate(-3,-29,V(-3,-29)),
 graph( 600, 600, -20, 20, -40, 40, (x + 3)^2 -29)) }}}


Part C: Determine the axis of symmetry for {{{f(t)}}}


Every parabola has an axis of symmetry which is the line that runs down its 'center'. This line divides the graph into two perfect halves.

In the picture of on the left, the axis of symmetry is the line {{{x = -3}}}.


{{{drawing( 600, 600, -20, 20, -40, 40,
circle(-3,-29,.2),locate(-3,-29,V(-3,-29)),line(-3,35,-3,-35),
 graph( 600, 600, -20, 20, -40, 40, (x + 3)^2 -29)) }}}