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Solution:
A general form o quadratic function is given by the following expression:
[tex]f(x)=ax^2+bx+c[/tex]The vertex form of this quadratic function is given by the following expression:
[tex]f(x)=a(x-h)^2+k[/tex]where (h,k) is the vertex of the quadratic function.
PART A:
To find the vertex form of the given function, we must complete the square of this function:
[tex]t^2+12t-18+6^2-6^2[/tex]this is equivalent to:
[tex](t+6)^2-18-6^2[/tex]that is:
[tex](t+6)^2-54[/tex]so that, the vertex form of the given function is:
[tex]f(t)=(t+6)^2-54[/tex]Part B: According to the previous part, we can conclude that the vertex is:
[tex](h,k)=(-6,-54)[/tex]Part C: The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola. So that, we can conclude that the axis of symmetry for f(t) is:
[tex]x\text{ =-6}[/tex]