Quadratic equation: factored and in vertex form

Consider the function f(x) = 2x² + 6x - 8

Part A.
What is the vertex form of f(x)?
A.)  f(x) = 2(x - 3)² - 4
B.)  f(x) = 2(x + 3)² + 4
C.)  f(x) = 2(x - 1.5)² - 12.5
D.)  f(x) = 2(x + 1.5)² - 12.5

Part B.
What is the factored form of f(x)?
A.)  f(x) = (2x + 1)(x - 8)
B.)  f(x) = (2x - 1)(x + 8)
C.)  f(x) = 2(x + 4)(x - 1)
D.)  f(x) = 2(x - 4)(x + 1)

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Part A.
Vertex form is f(x) = a(x - h)² + k
(h,k) are the (x,y) coordinates of the vertex

To find the x-coordinate of the vertex, find -b/2a

In f(x) = 2x² + 6x - 8,
a = 2, b = 6, and c = -8

-b
----
2a

=  -6
    ----
   2(2)

=  -6/4
x =  -1.5

To find the y-coordinate of the vertex,
substitute -1.5 back in for x.

y = 2(-1.5)² + 6(-1.5) - 8
=  2(2.25) - 9 - 8
=  5.5 - 9 - 8
=  -12.5

The vertex is (-1.5, -12.5).

Therefore, vertex form is
D.)  f(x) = -2(x + 1.5)² - 12.5

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Part B.
Factor f(x) = 2x² + 6x - 8
=  2(x² + 3x - 4)   {factored 2 out}
=  2(x + 4)(x - 1)  {factored into two binomials}

C.)  f(x) = 2(x + 4)(x - 1)