# How do you graph #y=sqrt(x-3)+2# and how does it compare to the parent function?

##### 1 Answer

Jun 13, 2018

See below

#### Explanation:

The second question answer the first, I'd say. Let's see the effect of the transformations: given a parent function

- Vertical stretch: we obtain it by multipling the whole function by some constant, so
#f(x) \to kf(x)# . If#0<k<1# we have a vertical shrink, if#k>1# we have a vertical expansion. If#k# is negative the function is reflected about the#x# axis, and then stretched as above. - Horizontal stretch: we obtain it by multiplying the variable by some constant, so
#f(x) \to f(kx)# . If#0<k<1# we have a horizontal expansion, if#k>1# we have a horizontal shrink. If#k# is negative the function is reflected about the#y# axis, and then stretched as above. - Vertical translation: we obtain it by adding some constant to the function, so
#f(x)\to f(x)+k# . If#k# is positive the shift is upwards, otherwise it's downwards. - Horizontal translation: we obtain it by adding some constant to the variable, so
#f(x)\to f(x+k)# . If#k# is positive the shift is leftwards, otherwise it's rightwards.

In your case, you have both vertical and horizontal translation, so you have

The first transformation translates the function three units to the right, the second one two units higher. See the graph of the three steps to check:

**Parent function:**

graph{sqrt(x) [-3.5, 10, -0.5, 5]}

**First translation:**

graph{sqrt(x-3) [-3.5, 10, -0.5, 5]}

**Second translation:**

graph{sqrt(x-3)+2 [-3.5, 10, -0.5, 5]}