<p>If f(x) = ax^2 + bx + c, how must a and b be related so that the graph of f(x-3) will be symmetric about the y-axis?</p>
<p>(A) a=b
(B) b=0, a is any real number
(C) b=3a
(D) b=6a
(E) a= b/9</p>
<p>I have no idea how to even approach this problem. I know that since f(x-3) is symmetric, that means f(x) is shifted to the left 3 units, but I don't know how to go from there. The explanation in my book didn't help at all. Thanks! :)</p>
<p>Why don't you just use a graphing calculator? If you're good enough at keying things in and you know how to insert/delete numbers, you could solve this problem in less than a minute.</p>
<p>You mean just try plugging in numbers? Is that the quickest way to do it?</p>
<p>Quick enough.
Numbers work, if you have no idea.</p>
<p>Or:
If f(x-3) symmetric ..., then f(x) vertex x-coordinate is ...?
And the formula for a parabola vertex x-coordinate is ...?
Then the equation is ...!</p>
<p>Ok... I think I get it... haha thanks.</p>
<p>Do you remember the equation for solving a quadratic of the form ax^2+bx+c?</p>
<p>It's (-b plus or minus the square root of (b^2-4ac)/2a</p>
<p>For the solution set to be equally symmetrical, what has to happen in this equation?--</p>
<p>the part that has -b/2a has to be zero </p>
<p>then the remaining part is a symmetric solution--since both the positive and negative solutions are equal that is the absolute values are equal.</p>
<p>So when is -b/2a always zero?--it's when b is zero and a is a real number (well not zero anyway).</p>
<p>Now let's think about why this is so--it's because the turning point for a quadratic is always when the slope of the first derivative line = 0. The first derivative of ax^2 +bx + c =0 is 2ax +b. Setting this equal to zero gives us x = -b/2a</p>
<p>Now let's do the same by substituting x-3 in for x.</p>
<p>f(x-3) = a(x-3)^2 + b(x-3) + c<br>
f(x-3) = a (x^2 - 6x +9) + bx -3b + c
f(x-3) = ax^2 - 6ax + 9a +bx -3b + c</p>
<p>The first derivative of this is 2ax - 6a + b or 2a (x-3) + b </p>
<p>For this to be symmetrical around the y-axis, the turning point has to be at
x =0 </p>
<p>So 2a (0-3) + b = 0 or b =0 and a is any real number (again!)</p>
<p>In other words it doesn't change just because it shifts left or right.</p>
<p>P.S. the answer b=3a is designed to confuse people, since this is the correct answer for when it is symmetrical around the x-axis--but that's a longer proof.</p>
<p>You know, I hate it when I make a mistake--and in this case, in my rush to get stuff ready for a meeting I have later today, I made a mistake on the last line of this calculation.</p>
<p>Everything is correct down to where I say: 2a (0-3) +b =0, then I go wrong</p>
<p>Obviously, if 2a (-3) + b =0 , then b = 6a (the correct answer)--sorry for any confusion.</p>