<ol>
<li>The graph in the xy-plane of the function f(x) = ax^2+bx+c is a parabola with a vertext at (3,-6). If a>0, which of the following cannot be a value of c.
a.6
b.3
c.0
d.-3
e.-6</li>
</ol>
<p>e. </p>
<p>Just imagine a parabola in your head. It is opening upwards (a>0), so the y-intercept (c) has to be greater than -6.</p>
<p>but doesn’t c have to be -6 since its in the form of (x-3)^2-6 ?</p>
<p>
</p>
<p>The form you mention is a(x-h)^2 + k, where the vertex is (-h,k)
The function in the question is in the form ax^2 + bx + c, where c is the y-intercept.</p>
<p>right and why can’t -6 be the y intercept?</p>
<p>Fact #1. Because a>0, the vertex, (3, -6) is the lowest point on the parabola. That is, the minimum value of the function is -6, and we know that this value occurs when x = -3.</p>
<p>Fact #2. The y-intercept of this graph is c. f(0) = a(0)^2 + b(0) + c. In other words, another point on this parabola is (0, c).</p>
<p>But, since (3, -6) is the lowest point on the parabola, c must be above -6.</p>
<p>AHHH i get it. basically, the parabola crosses the y axis at a higher point than -6… i see.</p>
<p>Yeah, I guess if I’d just said that, it would have been simpler!</p>
<p>haha thanks… i didn’t know the way i wrote it was in a different form from the ax^2 form. also, how do you get the vertex in the ax^2 form?</p>
<p>If you are given the form ax^2 + bx + c and want to find the vertex you simply plug it into:</p>
<p>
h = -b/2a
</p>
<p>Now that you have h, multiply it by -1 and plug it into ax^2 + bx + c to find k.</p>
<p>why multiply by -1?</p>
<p>and another question.</p>
<ol>
<li>If the range of the function y=f(x) is all real numbers between -1 and 12, inclusive, then what is the maximum value of g(x) if g(x) = 2f(x-1)+3</li>
</ol>
<ol>
<li>f(x-1) is just a translation of f, and the +3 is also just a translation. Translations don’t affect the range.</li>
</ol>
<p>Nspired: to convert from polynomial form (y = ax^2 + bx + c) to vertex form (y = a(x-h)^2 + k), you have to complete the square.</p>
<p>Greed: I disagree about the range. In Nspired’s example, only the horizontal shift (the -1) has no effect on the range. The coefficient of 2 is going to stretch the graph vertically by a factor of 2, thereby doubling the range, and the +3 is going to shift the graph upward 3, thereby raising both the lower and upper limits of the range by 3.</p>
<p>So, if the range of f was [-1, 12], the 2 will make the range [-2, 24] and the +3 will make it [1, 27]. The range of g is [1, 27].</p>
<p>ahhh i get it… its 2f(x) essentially doubling the range, then translated up 3 more. so 27… right?</p>
<p>Yes. Similarly, for the lower limit, 2(-1) + 3 = 1–not that this question asked for that information.</p>
<p>You are right. I don’t know what I was thinking…</p>
<p>Well, I wasn’t there, but I suspect you were thinking only about horizontal translations, which don’t affect the range.</p>
<p>(This is what math teachers do. We spend a lot of time trying to figure out what students were thinking. Grading an algebra test is a little bit like forensic mathematics.)</p>