Question

I only need help with this one.
Can you also explain it for me and how I can solve future problems like this?

Answer 1

horizontal
hmm

that means see only the left to right movement

ok, to move a function to the right c units, minus c from EVERY x
to move a function up c units, add c to the whole function


so we see the x terms
from x^2 to (x-4)^2+2
the +2 is irrelivant

4 was minused from every x
it was moved to the right 4 units


answer is 4


or you could do
vertex of f(x) is (0,0) and vertex of g(x) is (4,2) so from 0 to 4 is 4 to the right

answer is 4

Answer 2

$\bf \qquad \qquad \qquad \qquad \textit{function transformations} \\ \quad \\\\ \begin{array}{rllll} % left side templates f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ y=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}} \\ \quad \\ f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}} \\ \quad \\ f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}} \end{array}$

so hmm notice that template above, keep in mind that 

$\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative} \\\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \end{array}$

$\bf \begin{array}{llll} \bullet \textit{ vertical shift by }{{ D}}\\ \qquad if\ {{ D}}\textit{ is negative, downwards}\\\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{{{ B}}} \end{array}$

so hmm with that in mind, let's take a peek of yours

$\bf \begin{array}{llllll} g(x)=&(&x&-4)^2&+2\\\\ g(x)=&1(&1x&-4)^2&+2\\ &\uparrow &\uparrow &\uparrow &\uparrow \\ &A&B&C&D \end{array}$

C is -4, thus, the horizontal shift from f(x), is to the "right by 4 units", or  you can use the C/B notation, will will be -4/1 or -4 anyway