<p>page 490 #11
The quadratic function g is given by g(x)=ax^2+bx+c, where a and c are negative constants. which of the following could be the graph of g?</p>
<p>the answer is a graph that opens downwards and has a negative y-intercept. i plugged in a few different variables with -a and -c and managed to shift the graph to a positive x and y-intercept (equation is -x^2+6x-x). what confuses me is that there's an answer with a similar graph. is there some piece of information i'm failing to see? thanks!</p>
<p>if you compare expressions
ax^2+bx+c
and
-x^2+6x-x,
you'll see that there is no "c" in the second one (i.e. c=0): c should be <0, it's a constant.
For what x graph of any f(x) intercepts y axes (let set of all real numbers be a domain)?
Plug this value in a given function.</p>
<p>The answer is A. i don't exactly see what you are missing, BUT your equation of -x^2 + 6x -x doesn't correspond with ax^2 + bx + c, because c is a constant not a vairable....</p>
<p>So, as a rule, c will indicate if the graph moves vertically. So, -ax^2 + bx -c (the new equation) would only correspond with A.</p>
<p>plug in 0 into the equation,,,in A the g(0)= -1, but in C g(0) = 0, meaning in graph C, generally speaking, the graph does not have a negative constant c.</p>
<p>Bump. My friend just asked me this question, and amazingly, I think the SAT is wrong on this one. Let’s say we have a and c as -1 and -2, respectively. Then, make b something like 5, and you get a graph that looks like C. Plug in 0 for b, and you get something that looks like A. Plug in -5 for b, and you get something that looks like B. He said the book said the answer is A, and that’s what I would probably pick if it was a real test, but neither of us can understand why, because B and C are also valid.</p>
<p>Edit: just read the post above. I really feel like an idiot now. This wasn’t a particularly hard question, but pretty tricky. Thank you and sorry for bumping.</p>