A new type of question on the Digital SAT involves graph translations. Translations involve shifting the graph of a function in various directions within the coordinate plane. The two main types of translations we will focus on are vertical and horizontal translations. These questions can be tough and often deal with applying your knowledge about translations to other equation and graph question types!

To answer these questions, we have to understand the effects of translating a graph in the vertical and horizontal directions.

- A function (f(x)) can be vertically shifted upwards by adding a constant (k) to the function to get (f(x)+k)
- Similarly, a function (f(x)) can be vertically shifted downwards by subtracting a constant (k) to get (f(x)-k)

For example, if you have the function (f(x)), the graph of (y=f(x)-3) will be the graph of (f(x)) shifted down by 3 units.

The function in black shows the graph of the original equation (f(x)). The function in red shows the equation (f(x)-3) where the original function has undergone a vertical translation down three units. In a vertical translation, all of the *y*-values of the graph shift in the direction of the translation, indicated in how both the *y-*values of the *x* and *y*-intercepts shifted down by 3 units.

Horizontal translations work to shift the graph right or left.

- To shift the graph to the right, you subtract a constant (h) from the variable (x) to get (f(x-h))
- To shift the graph to the right, you add a constant (h) from the variable (x) to get (f(x+h))
- Notice how for horizontal translations, the constant is always INSIDE the function argument!

For example, the graph of (y=f(x-2)) will shift the graph of (y=f(x)) to units to the right.

The function in black shows the equation (f(x)), while the function in red shows the equation (f(x-2)). In a horizontal translation, all of the *x*-values of the function are shifted in the direction of the translation. You can see this in how the *x*-values of both of the graph’s intercepts shifted 2 units to the right on the new graph.

**Strategy****:**

**Understand how vertical and horizontal translations can change graphs and apply it to the question.****Use Desmos to graph equations when applicable.**

Let’s look at an example problem now and see if we understand what we have learned about translations.

The graph of (y=f(x)+7) is shown above. What equation defines function (f)?

A) (f(x)=-frac12x-7)

B) (f(x)=-frac12x-1)

C) (f(x)=-frac12x+6)

D) (f(x)=-frac12x+9)

We know that the graph depicts the function (f(x)+7) where (f(x)) has been shifted vertically up by 7 units. Thus, the graph of (f(x)) must be whatever we would get by shifting all the points on the graph back down by 7 units. If we shift the *y*-intercept of the graph down 7 units, our new *y*-intercept is at the point ((0, -1)). So we know that the graph of (f(x)) must have a *y*-intercept at -1. The only equation that matches this is B, so we know that B is the correct answer.

Here is another example that makes us apply our knowledge of translations.

The function (f) is defined by (f(x)=(x+3)(x-1)(x-4)). In the *xy*-plane, the graph of (y=h(x)) is the result of translating the graph of (y=f(x)) down 5 units. What is the value of (h(0))?

The first thing we should do to answer this question is to graph the function (f) in Desmos. We get the following graph:

From the graph, we can see that (f(0) is equal to 12. We are told that the graph of (y=h(x)) is the result of translating the graph of (y=f(x)) down 5 units. So to find the value of (h(0)), we shift the value of (f(0)) down 5.

(12-5=7) so we know that the value of (h(0)) is 7.

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