Question

transform the solid black function to match the dotted function.
+ ∨ f( + ∨ x+ 0 )+ 0

Explanation:

Step1: Analyze Vertical Shift

The dotted function is above the solid function. To shift up, we need to add a positive value to the function. Looking at the y - intercepts: solid has y - intercept around 0, dotted has y - intercept around 6. So vertical shift is +6.

Step2: Analyze Horizontal Shift

The solid function and dotted function: let's check the shape. The dotted function seems to be a horizontal shift? Wait, no, first vertical shift. Wait, the general form is \(a f(b(x - h))+k\), but here the form is \(f(\pm x + c)+d\). Wait, the given form is \(+ \ f(+ \ x + \square)+ \square\). Let's re - express. The standard transformation: \(f(x)\) to \(f(x - h)+k\), where \(h\) is horizontal shift (right if \(h>0\), left if \(h < 0\)) and \(k\) is vertical shift (up if \(k>0\), down if \(k < 0\)). In the given form \(f(+x + c)+d=f(x + c)+d\), which is equivalent to \(f(x-(-c))+d\). So horizontal shift is \(-c\) (left if \(c>0\), right if \(c < 0\)) and vertical shift is \(d\).

Looking at the graphs, the dotted function is a vertical shift up and maybe a horizontal shift? Wait, the solid function and dotted function: the dotted function is above the solid function. Let's check the y - values. The solid function at \(x = 0\) is around 0, the dotted function at \(x = 0\) is around 6. So vertical shift \(k = 6\). For the horizontal part, the form is \(f(+x+0)+6\)? Wait, no, maybe the horizontal shift? Wait, the given boxes: the first box (inside the function) and the second box (outside). Wait, the original form is \(+ \ f(+ \ x+\square)+ \square\). Let's assume the solid function is \(y = f(x)\), and the dotted function is \(y=f(x)+6\) (vertical shift up 6 units) and maybe no horizontal shift (so the first box is 0, second box is 6). Wait, let's re - check the graph. The solid function and dotted function: the dotted function is a vertical shift up. So the transformation is \(f(x)\) (solid) to \(f(x)+6\) (dotted), and no horizontal shift (so the first box is 0, second box is 6).

Wait, the initial form is \(+ \ f(+ \ x+\square)+ \square\). So the first “+” is for the sign of the function (reflection over x - axis? No, the dotted function is not a reflection, it's a shift). The second “+” is for the sign of the x - term (reflection over y - axis? No, the dotted function is not a reflection over y - axis). So the first “+” means no reflection over x - axis (\(a = 1\)), the second “+” means no reflection over y - axis (\(b = 1\)). Then the inside the function: \(x+\square\), and outside: \(+\square\).

So the vertical shift: the dotted function is above the solid function. Let's take a point on the solid function, say the vertex or a point. The solid function has a local maximum around \(x=-3\) with \(y = 3\), the dotted function has a local maximum around \(x=-3\) with \(y=3 + 6=9\)? Wait, no, the dotted function's local maximum is around \(y = 7\), solid's local maximum is around \(y = 3\). So the difference is 4? Wait, maybe my initial estimate was wrong. Wait, the dotted function at \(x = 0\) is around \(y = 6\), solid at \(x = 0\) is around \(y = 0\). So vertical shift of 6. And no horizontal shift (so the first box is 0, second box is 6).

So the transformation is \(+ \ f(+ \ x + 0)+6\).

Answer:

The first box (inside the function) is \(0\), the second box (outside the function) is \(6\). So the transformation is \(+ \ f(+ \ x+0)+6\).