Updated super hard math question?

Question

Consider the quadratic relation y=2x²+12x+54.

a) Use two different methods to determine the equation in vertex form.

b) Which method(s) enable you to also find points on the graph that are equidistant from the axis of symmetry?

c) Which method reveals the x-intercepts of the graph.

d) If the graph of the relation is not required, is there an advantage to using one method over another? Explain your reasoning.

Answer 1

Different methods for finding the vertex:
1) Complete the square
2) Use the formula. The vertex occurs at x = -b/(2a)
3) Use calculus

2x^2 + 12x + 54
2(x^2 + 6x) + 54
2(x^2 + 6x + 9 - 9) + 54
2(x^2 + 6x + 9) - 2(9) + 54
y = 2(x+3)^2 + 36
vertex: (-3,36)

a = 2, b = 12
x = -b/2a
x = -12/(2*2)
x = -3
y = 2(-3)^2 + 12(-3) + 54
y = 36
vertex: (-3,36)

(b) and (c) Method 1
(d) A lot of people have trouble completing the square, so it may be easier to use method 2.
If you've learned calculus, then method 3 is the easiest and quickest.

Answer 2

y-54 = 2(x^2 + 6x)

y - 54 + 18 = 2(x^2 + 6x + 9)

y - 36 = 2(x+3)^2 *************

Your vertex, then, is (-3,36)

I'll let someone else tackle b,c,d.

Another way to find the vertex is to use the rule that the x-coord of the vertex = -b/2a. So you get -12/2*2=-12/4 = -3 for x. To find y, you plug in -3 for x back in the original equation
y = 2(9)+12(-3)+54 = 18 - 36 + 54 = =36. So then (-3, 36) is your vertex.

Putting this into the eqn. y-k = a(x-h)^2, we get
y-36 = 2(x+3)^2 *************