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For each of the functions below, state the domain and range, the restrictions, the intervals of increasing and decreasing, the roots, y-intercepts, and vertices.

1) f(x)=2x^28

2) f(x)=+x - 2

3) f(x)=x+1/x-1 
       

 Jul 12, 2015

Best Answer 

 #1
avatar+118696 
+5

ok I will take a look  :))

For each of the functions below, state the domain and range, the restrictions, the intervals of increasing and decreasing, the roots, y-intercepts, and vertices.

1) f(x)=2x^2−8

2) f(x)=+√x - 2

3) f(x)=x+1/x-1 

 

1)f(x)=2x28

  Any real number can be squared so x can be any real number  

             Domain:    xR        (x is in the set of reals)     (,)

  x2 cannot be negative,   it can be any positive number though.  Same for 2x2

 2x28      cannot be any smaller than     -8

             Range:           8f(x)<[8,)

 

Even before you went to all this bother you should have recognised that this is a concave up parabola.

It is a steeper version of y=x^2   And every point is dropped 8 units.

In other words,   The axis of symmetry is x=0     (That is the y axis)

The vertex is (0,-8)

The y intercept is -8

It is decreasing when x

It is increasing when x>0  (the tangent has a positive gradient)

When f(x)=0      

0=2x280=x240=(x2)(x+2)x2=0orx+2=0x=2orx=2$Therootsarex=2andx=2$

 

I think that covers the first one.

Think about it and aske questions if you have any.

Have a go at the second one yourself and present what you think.  

 

Here is the graph:

https://www.desmos.com/calculator/gmxuxxwtbv

 

I will talk about it with you then :) 

 Jul 12, 2015
 #1
avatar+118696 
+5
Best Answer

ok I will take a look  :))

For each of the functions below, state the domain and range, the restrictions, the intervals of increasing and decreasing, the roots, y-intercepts, and vertices.

1) f(x)=2x^2−8

2) f(x)=+√x - 2

3) f(x)=x+1/x-1 

 

1)f(x)=2x28

  Any real number can be squared so x can be any real number  

             Domain:    xR        (x is in the set of reals)     (,)

  x2 cannot be negative,   it can be any positive number though.  Same for 2x2

 2x28      cannot be any smaller than     -8

             Range:           8f(x)<[8,)

 

Even before you went to all this bother you should have recognised that this is a concave up parabola.

It is a steeper version of y=x^2   And every point is dropped 8 units.

In other words,   The axis of symmetry is x=0     (That is the y axis)

The vertex is (0,-8)

The y intercept is -8

It is decreasing when x

It is increasing when x>0  (the tangent has a positive gradient)

When f(x)=0      

0=2x280=x240=(x2)(x+2)x2=0orx+2=0x=2orx=2$Therootsarex=2andx=2$

 

I think that covers the first one.

Think about it and aske questions if you have any.

Have a go at the second one yourself and present what you think.  

 

Here is the graph:

https://www.desmos.com/calculator/gmxuxxwtbv

 

I will talk about it with you then :) 

Melody Jul 12, 2015
 #2
avatar
0

I tried the other two questions but still dont understand where to start because the equation doesnt look the same as the first question. What do i do? The square root and divison between the x's terrifies me :(

 Jul 12, 2015

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