First, we need to get this function into a more standard form.

$\displaystyle g(x)=4(2x-6)^{5}$

$\displaystyle g(x)=4[2(x-3)]^{5}$

Now we can see that while the function is being horizontally compressed by a factor of 2, it's being translated 3 units to the right, not 6. (It's also being vertically stretched by a factor of 4, of course.)

By definition, $\displaystyle (g \circ f) (x) = g (f(x))$, so

$\displaystyle (g \circ f) (x) = g ( 6x-7 )$

$\displaystyle = 2 (6x-7) + 8$

$\displaystyle =12 x-14 + 8$

$\displaystyle =12 x - 6$

By definition, $\displaystyle (g \circ f) (x) = g (f(x))$, so

$\displaystyle (g \circ f) (x) = g (2x+7)$

$\displaystyle = \left ( 2x+7\right )^{2} - 5$

$\displaystyle = \left ( 2x\right )^{2} + 2 \cdot 2x \cdot 7 +7 ^{2} - 5$

$\displaystyle =4 x ^{2} + 28 x +49 - 5$

$\displaystyle =4 x ^{2} + 28 x +44$

To transform the function horizontally, we must make an addition or subtraction to the input, x. Because we are asked to move the function to the left, we must add the number of units we are moving. This is the opposite of what one would expect, but if we are inputting values that are to the left of the original, they are less than what would have originally been. So, to counterbalance this, we add the units of the transformation.

For our function being transformed five units to the left, we get

$\displaystyle f(x)=(x+5)^2+5$

To transform a function horizontally, we must add or subtract the units we transform to x directly. To move left, we add units to x, which is opposite what one thinks should happen, but keep in mind that to move left is to be more negative. To flip a function, the entire function changes in sign.

After making both of these changes, we get

$\displaystyle -[(x+1)^2+4]$

To transform a function we use the following formula,

$\displaystyle f(x)=(x-h)^2+v$

where h represents the horizontal shift and v represents the vertical shift.

In this particular case we want to shift to the left five units,

$\displaystyle h=-5$

and vertically up two units,

$\displaystyle v=2$.

Therefore, the transformed function becomes,

$\displaystyle \\f(x)=(x-h)^2+v \\f(x)=(x--5)^2-1+2 \\f(x)=(x+5)^2+1$.

Expand the binomial.

$\displaystyle (x+1)(x+1) = x^2+2x+1$

Multiply negative by this quantity.

$\displaystyle -2 (x^2+2x+1) = -2x^2-4x-2$

The polynomial in standard form is:  $\displaystyle y = -2x^2-4x-2$

Shifting this graph up one will change the y-intercept by adding one unit.

The answer is:  $\displaystyle y = -2x^2-4x-1$

Shifting this parabola up two units requires expanding the binomial.

Use the FOIL method to simplify this equation.

$\displaystyle y= (6-3x)^2 = (6-3x)(6-3x)$

$\displaystyle y=(6)(6)+(6)(-3x)+(-3x)(6)+(-3x)(-3x)$

$\displaystyle y=36-18x-18x+9x^2$

$\displaystyle y=9x^2-36x+36$

Shifting this graph up two units will add two to the y-intercept.

The answer is:  $\displaystyle y=9x^2-36x+38$

We will need to determine the equation of the parabola in standard form, which is:

$\displaystyle y=ax^2+bx+c$

Use the FOIL method to expand the binomials.

$\displaystyle (x-4)(x+1) = x^2+x-4x-4 = x^2-3x-4$

Shifting this up two units will add two to the value of $\displaystyle c$.

The answer is:  $\displaystyle y=x^2-3x-2$

In this problem, the $\displaystyle x$ in the $\displaystyle g(x)$ equation becomes $\displaystyle f(x)$ --> $\displaystyle g((2x-5)^{}2-1)$.

This simplifies to $\displaystyle 4x^{}2-20x+25-1$, or

$\displaystyle 4x^{}2-20x+24$.