Which absolute value function has a graph that is wider than the parent function, f(x) = |x|, and is translated to the right 2 u
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Which absolute value function has a graph that is wider than the parent function, f(x) = |x|, and is translated to the right 2 units? f(x) = 1.3|x| – 2f(x) = 3|x – 2|f(x) = 3/4 |x – 2|f(x) = 4/3 |x| + 2
Which absolute value function has a graph that is wider than the parent function, f(x) = |x|, and is translated to the right 2 units? f(x) = 1.3|x| – 2f(x) = 3|x – 2|f(x) = 3/4 |x – 2|f(x) = 4/3 |x| + 2
2 answers:
Answer:
Required function -
Step-by-step explanation:
Given : The parent function and is translated to the right 2 units.
To find : Which absolute value function has a graph that is wider than the parent function?
Solution :
The parent function
with the vertex (0,0)
The parent function is translated to the right 2 units.
Transformation to the right,
f(x)→f(x-b) , the graph of f(x) is shifted towards right by b unit.
Same as the graph f(x) is shifted towards right by 2 unit and form graph of h(x).
If the graph is wider than the parent function then the function must be in the form of,
Where the value of k must be less than of equal to 1. If k is more than 1 then the graph compressed.
So, let it be k=1
Therefore, The required absolute value function is
We plot the graph of both the equations in which translation is shown.
Refer the attached graph below.
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Answer:
- x = ±√3, and they are actual solutions
- x = 3, but it is an extraneous solution
Step-by-step explanation:
The method often recommended for solving an equation of this sort is to multiply by the product of the denominators, then solve the resulting polynomial equation. When you do that, you get ...
... x^2(6x -18) = (2x -6)(9)
... 6x^2(x -3) -18(x -3) = 0
...6(x -3)(x^2 -3) = 0
... x = 3, x = ±√3
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Alternatively, you can subtract the right side of the equation and collect terms to get ...
... x^2/(2(x -3)) - 9/(6(x -3)) = 0
... (1/2)(x^2 -3)/(x -3) = 0
Here, the solution will be values of x that make the numerator zero:
... x = ±√3
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So, the actual solutions are x = ±3, and x = 3 is an extraneous solution. The value x=3 is actually excluded from the domain of the original equation, because the equation is undefined at that point.
_____
<em>Comment on the graph</em>
For the graph, we have rewritten the equation so it is of the form f(x)=0. The graphing program is able to highlight zero crossings, so this is a convenient form. When the equation is multiplied as described above, the resulting cubic has an extra zero-crossing at x=3 (blue curve). This is the extraneous solution.
