Method 1:

The x-intercepts are \(\displaystyle x = -6, 2\).  These values would be obtained if the original quadratic were factored, or reverse-FOILed and the factors were set equal to zero.

For \(\displaystyle x = -6\), \(\displaystyle x + 6 = 0\).  For \(\displaystyle x = -2\), \(\displaystyle x + 2 = 0\).  These equations determine the resulting factors and the resulting function; \(\displaystyle f(x) = (x + 6)(x + 2)\).

Multiplying the factors and simplifying, 

\(\displaystyle f(x) = x\cdotx + 2\cdotx + 6\cdotx + 6\cdot2 = x^{2} + 8x + 12\).

Answer: \(\displaystyle f(x) = x^{2} + 8x + 12\).

Method 2:

Use the form \(\displaystyle (x - h)^{2} + k\), where \(\displaystyle (h, k)\) is the vertex. 

\(\displaystyle (h, k)\) is \(\displaystyle (-4, -4)\), so \(\displaystyle h = -4\), \(\displaystyle k = -4\).

\(\displaystyle (x - [-4])^{2} + (-4) = (x + 4)^{2} - 4\)

Answer: \(\displaystyle f(x) = (x + 4)^{2} - 4\)

                       \(\displaystyle = (x + 4)(x + 4) - 4\)

                       \(\displaystyle = x^{2} + 8x +16 - 4\)

                       \(\displaystyle = x^{2} + 8x + 12\)

The zeros for this polynomial are \(\displaystyle -1.5, \frac{1}{3}, 4\).

This means that the factors are equal to zero when these values are plugged in for x.

\(\displaystyle x + 1.5 = 0\) multiply both sides by 2

\(\displaystyle 2 x + 3 = 0\) so one factor is \(\displaystyle 2x+3\)

\(\displaystyle x - \frac{1}{3} = 0\) multiply both sides by 3

\(\displaystyle 3x - 1 = 0\) so one factor is \(\displaystyle 3x-1\)

\(\displaystyle x - 4 = 0\) so one factor is \(\displaystyle x- 4\)

Multiply these three factors:

\(\displaystyle (2x+3) ( 3x-1) = 6x^2 + 9 x - 2 x - 3 = 6x^2 + 7 x - 3\)

\(\displaystyle (6x^2 + 7 x - 3 ) ( x - 4 ) = 6x^3 + 7x^2 - 3 x \enspace -24x^2 - 28x + 12\)\(\displaystyle = 6x^3 - 17x^2 -31x+12\)

The zeros of this polynomial are \(\displaystyle 4, -4 , -1.5\). This means that the factors equal zero when these values are plugged in.

One factor is \(\displaystyle x - 4\)

One factor is \(\displaystyle x + 4\)

The third factor is equivalent to \(\displaystyle x + 1.5\). Set equal to 0 and multiply by 2:

\(\displaystyle 2(x + 1.5) = 2(0 )\)

\(\displaystyle 2x+3 = 0\)

Multiply these three factors:

\(\displaystyle (x-4)(x+4) = x^2 - 16\)

\(\displaystyle (x^2 - 16) (2x+3 ) = 2x^3 +3x^2 - 32x - 48\)

The graph is negative since it goes down then up then down, so we have to switch all of the signs:

\(\displaystyle y = -2x^3 -3x^2 + 32x + 48\)

The zeros of the polynomial are \(\displaystyle 0.6, 3, -2\). That means that the factors equal zero when these values are plugged in.

The first factor is \(\displaystyle x - 0.6 = 0\) or equivalently \(\displaystyle x - \frac{3}{5} = 0\) multiply both sides by 5: 

\(\displaystyle 5x- 3 = 0\)

The second and third factors are \(\displaystyle x - 3 = 0\) and \(\displaystyle x + 2 = 0\)

Multiply:

\(\displaystyle (5x - 3 ) ( x - 3 ) = 5x^2 -3x -15x +9 = 5x^2 -18x + 9\)

\(\displaystyle (5x^2 -18x + 9 ) ( x + 2 ) = 5x^3 -18x^2 +9x \enspace + 10x^2 -36x +18\)

\(\displaystyle = 5x^3 - 8x^2 -27x + 18\)

Because the graph goes down-up-down instead of the standard up-down-up, the graph is negative, so change all of the signs:

\(\displaystyle -5x^3 +8x^2 +27x - 18\)

The zeros for this polynomial are \(\displaystyle -1.75, \frac{1}{3}, 4\). That means that the factors are equal to zero when these values are plugged in. 

\(\displaystyle x + 1.75 = 0\) or equivalently \(\displaystyle x + \frac{7}{4} = 0\) multiply both sides by 4

\(\displaystyle 4x+7 = 0\) the first factor is \(\displaystyle 4x+7\)

\(\displaystyle x-\frac{1}{3} = 0\) multiply both sides by 3

\(\displaystyle 3x-1 = 0\) the second factor is \(\displaystyle 3x - 1\)

\(\displaystyle x - 4 = 0\) the third factor is \(\displaystyle x - 4\)

Multiply the three factors:

\(\displaystyle (4x+7)(3x-1) = 12x^2 +21x -4x - 7 = 12x^2+ 17x - 7\)

\(\displaystyle (12x^2 +17x - 7 )(x-4) = 12x^3 +17x^2 -7x + \enspace -48x^2 -68x +28\)

\(\displaystyle = 12x^3 -31x^2 -75x +28\)