Questions from Barron's Math IIC Tests

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1

<p>I need some help with understanding these questions, if you please.</p>

<p>To indicate squares I have multiplied the squared expression to itself for convenience.</p>

<p>First one's coming from Model Test 1, q.47 page 150. It says:</p>

<p>If f(x)=3x.x+4x+5, what will be the value of k equal so that the graph of f(x-k) will be symmetric to the y-axis?</p>

<p>(A) 2/3 (B)-2/3 (C)0 (D)-4 (E)-4/3</p>

<p>The answer is A, and the only solution the book gives guides me through a solution involving a graphing calculator.</p>

<p>The other one is from Model Test 2, the last question, q.50 page 170. It says:</p>

<p>If (x,y) represents a point on the graph of y=2x+1, which of the following could be a portion of the graph of the set of points (x,y.y)?</p>

<p>The choices are a bunch of graphs.</p>

<p>Normally i graphed y=2x+1 and and using the same x values I found the new y values(for y.y). Anyway, the correct choice is C, it reads:</p>

<p>Note that the graph of y=(2x+1)(2x+1)=4(x+1/2)(x+1/2) is a parabola with vertec at (-1/2,0) that opens up. The only answer choice with these properties is C.</p>

<p>What I don't get from this solution is the reason why the book squares 2x+1 and not y.</p>

<p>Any help is appriciated. Thanks for your time.</p>

2

<p>for the second one, the function is being squared.</p>

3

<p>Ill take a crack at the first one.</p>

<p>Okay, so f(x) is a parabola. In order for a parabola to be symmetric over the y axis, it has to have its vertex along it. At its vertex, the slop always = 0. So what I did was take the derivative of the function, which gives [f'(x) = 6x + 4]. when plugging in f'(x-k), you get [f'(x-k) = 6x -6k + 4]
Now at the vertex, x = 0, and f'(x) = 0. So 0 = 0 - 6k + 4, or 6k = 4, or k = 2/3. </p>

<p>As for w/o doing derivatives, I would imagine you would have to solve for the standard form. </p>

<p>While doing that it goes something like:</p>

<p>y = 3xx + 4x + 5
y - 5 = 3xx + 4x
y - 5 = 3(xx+4x/3+4/9) - 4/3
y - 11/3 = 3(x + 2/3)^2</p>

<p>so x would have to be switched with x - 2/3 in order for for the x cord of the vertex to be at x = 0.</p>

<p>As for the 2nd question, I am having a hard to understanding for what you are asking for.</p>

4

<p>I had previously used some differentiation when dealing with indeterminate forms in limits, but I can't pretend to be knowledgeable about the way you used it in your solution. Anyway, as far as my slack perception went, you used derivative to solve for k, by equating slope at f'(x-k) to f'(x-k).
I can't understand though what exactly can I find by differentiating a second degree equation. Albeit I am not sure, I think,

should be f'(x-k)=0.
If I am wrong, please be so kind to explain your steps clearly.</p>

<p>Regarding,

I was asking why the book in its solution squares 2x+1 instead of y, when the question blatantly asks for (x,y^2)</p>

<p>Thanks for your time and help.</p>

5

<p>"should be f'(x-k)=0.
If I am wrong, please be so kind to explain your steps clearly."</p>

<p>all I was trying to say was that the slope should be zero. It could possibly be better said that since x is supposed to equal zero, and f'(x-k) is supposed to equal zero, then f'(-k) is supposed to equal zero. If you plug that into the derivative you get the same result of 2/3</p>

6

<p>As for the 2nd one since y is the dependent variable, the corrdiante system can be written x,y = (x, 2x + 1), so x, y^2 = (x, (2x+1)^2). I would normally use t's but that would add lots of extra explaining.</p>

7

<p>2nd one's OK, thanks!</p>

<p>Regarding the first question, from your first post,

At its vertex the slope is always zero. Does a single point have a slope?
And my other question was: What exactly can i find by taking derivative of a function? I mean why would somebody differentiate a 2nd degree equation, generally speaking?</p>

8

<p>I don't know if this is what you're looking for but I graphed it when I did your first problem (#47). If something is symmetric to the y-axis that means when you fold it hotdog style it matches up. It's a parabola so you just have to put the vertex on the y-axis. The current vertex is (-2/3, y) (I found the minimum on my calc) and to move something left or right is counterintuitive and involves the x-value. Currently it is 2/3 units left of the y-axis. You want to move it right, so you subtract from your original x-value.</p>

<p>(x-K) means K already is being substracted, so choose the positive 2/3 as your answer.</p>

<p>I don't know if that made any sense, but I didn't use many mathematical terms so that could help.</p>

9

<p>Hey Seiken, I apologize for my incomprehension, but it was just a couple of hours ago when I got to know that differentiating an equation and substituting for x, gives the slope of the tangent line to that point.</p>

<p>Everything is clear now. Thank you!</p>

<p>ps. MissBarbara, thanks for sharing your unorthodox approach.</p>

10

<p>Actually missBarbara's approach is close to the standard approach.</p>

<p>for f(x) = a<em>x</em>x + b* x + c</p>

<p>the axis of symmetry is x = -b/2a, so it is x = -3/2</p>

<p>then you want to move the axis 3/2 unit to the right to coincide with the y axis, so you have f(x - 3/2) and that means k = 3/2.</p>