I need help with this 50 points
2 answers:
Answer:
Step-by-step explanation:
Vertically stretched. The action of vertically stretched is accomplished by altering a in
y = a* abs(x)
What that means is that you make a > 1. In this case, a = 2
So far, what you have is
y = 2*abs(x)
Six units down. The action of 6 units down is accomplished by a number added or subtracted to/from absolute(x). down is minus, up is plus.
y = 2*abs(x) - b. Since we are moving down, b<0
y = 2*abs(x) - 6
Four Units Right. This is the tough one because it is anti intuitive. You would think you should be adding something somewhere to get a right hand movement.
Not true.
To move right you subtract something in the brackets.
y = 2*abs(x - 4) - 6
Graph
Just to make things complete, I have graphed this for you. Desmos is wonderful for this kind of problem.
red: y = abs(x)
blue: y = 2*abs(x - 4) - 6
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So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:
Next, square both sides:
Next, subtract x and add 2 to both sides of the equation:
Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:
Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:
Now you can rewrite the equation as
Now, apply the Zero Product Property and solve for x as such:
Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:
Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>