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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: TylerD9 on April 15, 2019, 12:13:02 pm

Title: Modulus Question
Post by: TylerD9 on April 15, 2019, 12:13:02 pm: 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 :)
Title: Re: Modulus Question
Post by: Srd2000 on April 15, 2019, 12:30:06 pm: 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 :D
Title: Re: Modulus Question
Post by: TylerD9 on April 15, 2019, 04:24:42 pm: Quote from: Srd2000 on April 15, 2019, 12:30:06 pm
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 :D

Makes sense, thank you heaps !
Title: Re: Modulus Question
Post by: schoolstudent115 on April 16, 2019, 02:53:32 pm: (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:
$|n| = \sqrt{n^2}$
Substituing in the given function values:
(1) $f(3)=3 \implies |3-a|+b=3$
(2) $f(-1)=3 \implies |-1-a|+b=3$

Setting equal to each other (since they both equal 3):
$|3-a|=|-1-a|$
$\sqrt{(3-a)^2}=\sqrt{(-1-a)^2}$
$\sqrt{9-6a+a^2}=\sqrt{1+2a+a^2}$
$9-6a+a^2=1+2a+a^2 \implies 8-8a=0 \implies a=1$
Substitute 'a' into the first equation: $|3-1|+b=3 \implies 2+b=3 \implies b=1$

So $a=1, b=1$

The graph is: $f(x)=|x-1|+1$
The image of the graph is attached.
Title: Re: Modulus Question
Post by: TylerD9 on April 16, 2019, 07:09:09 pm: Quote from: schoolstudent115 on April 16, 2019, 02:53:32 pm
(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:
$|n| = \sqrt{n^2}$
Substituing in the given function values:
(1) $f(3)=3 \implies |3-a|+b=3$
(2) $f(-1)=3 \implies |-1-a|+b=3$

Setting equal to each other (since they both equal 3):
$|3-a|=|-1-a|$
$\sqrt{(3-a)^2}=\sqrt{(-1-a)^2}$
$\sqrt{9-6a+a^2}=\sqrt{1+2a+a^2}$
$9-6a+a^2=1+2a+a^2 \implies 8-8a=0 \implies a=1$
Substitute 'a' into the first equation: $|3-1|+b=3 \implies 2+b=3 \implies b=1$

So $a=1, b=1$

The graph is: $f(x)=|x-1|+1$
The image of the graph is attached.

Thank you :)