Modulus Question

[TylerD9]

Hello,

I am stuck on this question:
if f(x) = |x-a| + b with f(3)=3 and f(-1)=3, find a and b
Could someone please help me out?

Thank you

[Srd2000]

Hey Tyler, I may be able to help.

So we know that a modulus graph of a linear equation will be a big, symmetric V shape from the origin if we ignore the a and b things. Importantly, we're given that f(3) = f(-1) = 3 , therefore we must have a symmetry between them. Average those x values.
(3+ -1)/2 = 1    This means that our graph is shifted 1 across right. Thus, a = 1
Now we're just left with ol' b. How do we find that? Simple, we have f(x) = |x-1|+b. Substitute a point in a solve for b. b = 1

f(x) = |x-1|+1

Let me know if this doesn't make sense or I'm wrong. Good luck

[TylerD9]

Hey Tyler, I may be able to help.

So we know that a modulus graph of a linear equation will be a big, symmetric V shape from the origin if we ignore the a and b things. Importantly, we're given that f(3) = f(-1) = 3 , therefore we must have a symmetry between them. Average those x values.
(3+ -1)/2 = 1    This means that our graph is shifted 1 across right. Thus, a = 1
Now we're just left with ol' b. How do we find that? Simple, we have f(x) = |x-1|+b. Substitute a point in a solve for b. b = 1

f(x) = |x-1|+1

Let me know if this doesn't make sense or I'm wrong. Good luck

Makes sense, thank you heaps !

[schoolstudent115]

(I'm in year 10, not in spec yet so I'm not sure how exactly they want you to do it)
A different approach using the definition of the absolute value:
 
Substituing in the given function values:
(1)
(2)

Setting equal to each other (since they both equal 3):




Substitute 'a' into the first equation:

So

The graph is:
The image of the graph is attached.

[TylerD9]

(I'm in year 10, not in spec yet so I'm not sure how exactly they want you to do it)
A different approach using the definition of the absolute value:
 
Substituing in the given function values:
(1)
(2)

Setting equal to each other (since they both equal 3):




Substitute 'a' into the first equation:

So

The graph is:
The image of the graph is attached.

Thank you