Question

The function f(t) = t2 + 12t − 18 represents a parabola. Part A: Rewrite the function in vertex form by completing the square. Show your work. (6 points) Part B: Determine the vertex and indicate whether it is a maximum or a minimum on the graph. How do you know? (2 points) Part C: Determine the axis of symmetry for f(t). (2 points)

Answer 1

Complete the square:
f(t) = t² + 12t - 18, f(t) = (t² +12t) -18. 
Complete the square of t² +12t ↔ t² + 12t + 36 -36 = (t+6)² -36, ten:

f(t) = t² + 12t - 18 = (t+6)² -36 - 18 ↔f(t) = (t + 6)² - 54
f(t) = y = t + 6)² - 54 or ↔ y+54 = (t+6)²
(y+54) = (t+6)² is the standard form of a parabola. [(y-k) = a(x-h)²]
The vertex is (-54,-6) and since a is positive, te parabola opens upward with a minimum (-54,-6)
For the axis of symmetry, let's take again the initial form y= t²+12t-18.
(ax²+bx+c). The axis of symmetry is - b/2a = (-12)/2 = -6.
Then the axis of symmetry is x= -6

Answer 2

Part A:    
f ( t ) = t² + 12 t - 19
( t² + 12 t + 36 ) - 36 - 18
( t + 6)² - 54

Part B : The vertex ( - 6, -54 ).
This is the minimum, because the parabola is open up ( a > 0 ).

Part C : The axis of symmetry t = - 6.