<p>Q)
f(x)= | 3x-17 |
For the func defined above, what is one possible value of 'a' for which f(a) < a ?</p>
<p>Q)
In a model function question, the parabolas equation is y= ax^2 + 2; where a is a constt.
If y= (a/3)x^2 + 2
Which of the following describes the resulting graph compared to obv mentioned graph equation.
1) It will be narrower ( what i quess is right)
2) it will be wider
3) it will move to left
4) it will move to right
5) moved down 3 units.</p>
<p>1: 8. Simply plugging in some obvious numbers will get you the answer in about 10 seconds.
2: B. (or 2. in this case). Just imagine a graph with a = 6. If you use the former equation, you’ll end up with y much higher up for each and every x. If you make that ‘a’ smaller with the latter equation, you lower the place a y will be on the graph and make it look wider as the ‘x’ goes on in either direction (negative/positive).</p>
<p>I didnt get how you did the 1st one? How is f(a)< a related to the equation?</p>
<h1>1</h1>
<p>They give you a new variable just to confuse you. Stick with ‘x’ instead of ‘a’. If f(x)<x, then="" |3x-17|<x.="" removing="" the="" absolute="" value="" signs="" give’s="" you="" two="" separate="" equations:="" 3x-17<x="" and="" 3x-17="">-x. Solving for x in both equations gives you x<8.5 and x>4.25. Therefore, the answer is any number in the interval 4.5<x<8.5.</x,></p>
<h1>2</h1>
<p>This one doesn’t require any math at all. Intuition should tell you that in the equation ax^2+bx+c, if a=1, the graph is normal, if a<1, it is wider and if a>1, it is narrower. In the question, you are given that the equation’s leading coefficient is a/3, so the graph would be wider than when the leading coefficient is a.</p>
<ol>
<li><p>Try something like a=6. Then f(6) = |3(6) - 17| = 1, which is clearly less than 6.</p></li>
<li><p>B, this transformation makes it wider. If you replace a with 3a, for example, the graph will look “steeper.”</p></li>
</ol>