How do you sketch the graph of #y=x^2-5# and describe the transformation?

1 Answer
KillerBunny
May 10, 2018

See below

Explanation:

If you know the graph of a function #y=f(x)#, then you can have four kind of transformations: the most general expression is

#A f(wx+h)+v#

where:

  • #A# multiplies the whole function, thus stretching it vertically (expansion if if #|A|>1#, contraction otherwise)
  • #w# multiplies the input variable, thus stretching it horizontally (expansion if if #|w|<1#, contraction otherwise)
  • #h# and #v# are, respectively, horizontal and vertical translations.

In your case, starting from #f(x)=x^2#, you have #A=1# and #w=1#. Being multiplicative factors, they have non effect.

Moreover, #h=0#. Being ad additive factor, it has non effect.

Finally, you have #v = -5#. This means that, if you start from the "standard" parabola #f(x)=x^2#, the graph of #f(x)=x^2-5# is the same, just translated #5# units down.

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