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 #f(x)#, we have four basic transformation: stretched and translation, both horizontal and vertical.

  • 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

#sqrt(x) \to sqrt(x-3) \to sqrt(x-3)+2#

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: #y=sqrt(x)#
graph{sqrt(x) [-3.5, 10, -0.5, 5]}

First translation: #y = sqrt(x-3)#
graph{sqrt(x-3) [-3.5, 10, -0.5, 5]}

Second translation: #y = sqrt(x-3)+2#
graph{sqrt(x-3)+2 [-3.5, 10, -0.5, 5]}