Curves, relationships, functions and symmetry

  • #1
For the following curves i) y=x^2+4x-1 ii) y=-2+or-Square root(x+5)
a) Sketch both the curves on the same sheet of graph paper- against the same axis
I have done this, although I have not shown it here

b) Determine with proof, whether the above curves are related.
Not sure how to do this.

c)Determine with proof, if either of the above curves are a function
Not sure how to do this either.

d)Determine, with proof, whether either of the curves displays odd or even symmetry
y=f(x)=x^2+4x-1
f(-x)=-x^2+4x-1
Even function
Is this correct

Any help would be great thanks
 

Answers and Replies

  • #2
35,405
7,275
For the following curves i) y=x^2+4x-1 ii) y=-2+or-Square root(x+5)
a) Sketch both the curves on the same sheet of graph paper- against the same axis
I have done this, although I have not shown it here

b) Determine with proof, whether the above curves are related.
Not sure how to do this.
What did you notice about the two graphs in part a?
c)Determine with proof, if either of the above curves are a function
Not sure how to do this either.
Surely you have learned how to tell whether a graph represents a function. Does the phrase "vertical line test" ring a bell?
d)Determine, with proof, whether either of the curves displays odd or even symmetry
y=f(x)=x^2+4x-1
f(-x)=-x^2+4x-1
Even function
Is this correct
A function is even if f(-x) = f(x) for all x in the domain of the function. A function is odd if -f(-x) = f(x) for all x in the domain of the function. You have not calculated f(-x) correctly. Please try again.
Any help would be great thanks
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966
If [itex]y= 2\pm\sqrt{x+ 5}[/itex] then [itex]y- 2= \pm\sqrt{x+ 5}[/itex] and, squaring both sides, [itex](y- 2)^2= y^2- 4y+ 4= x+ 5[/itex] which is exactly the same as [itex]y^2- 4y- 1= x[/itex]. How is that formula connected to [itex]y= x^2- 4y+ 1[/itex]?
 

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