<p>Anyone who owns the bb here... please help! :(
Okay so the questions on p652 #8 and on p669 #6 I don't even know what the question is asking for... I am clueless!!!</p>
<p>On p652 #8 I thought it would be D but the answer says C.... I don't get it at all.
On p669 #6 I thought it would be C but the answer says B... I don't get this one AT ALL. Like What is it even giving me from the graph?</p>
<p>I find that many people find the function problems daunting. You just have to breathe and carefully analyze. Now the question seems like the other math problems.</p>
<p>Page 652 #8
The information is given that the function has a minimum at the point (1,1).
→ This tells us that x = 1 is the line of symmetry for this function. Also, the increments are 1 each.</p>
<p>If f(b) = f(3), what can b be?
→ Look at the graph and find f(3). You’ll find that f(3) is 5. For what other value of x does the y value equal 5? -1.</p>
<p>Or, you can approach the problem this way:
Since x = 1 is the line of symmetry, any y value for x = 1 + a is equal to x = 1 - a. For example, y value at x = 2 must equal the y value at x = 0.</p>
<p>f(3) must equal f(-1).</p>
<p>Page 669 #6
You are given that g(k) = 1.
→ Ask yourself, “when does the y value equal 1 in the graph of the function g?”
y value equals 1 when -1 ≤ x ≤ 0.
The only answer that satisfies this requirement is B.</p>
<p>Ok I get the P669 question. Man I was confused… I never really gave that one a thought… I got hard questions right but its those medium questions that ALWAYS trip me! </p>
<p>Correct me if I am wrong for #8 on p652…</p>
<p>Here is my way since the lowest point is 1,1… the axis of sym has to be 1</p>
<p>I can assume that -b/2a </p>
<p>so 1= -b</p>
<p>-b = -1</p>
<p>therefore 1 = -(-1)</p>
<p>
</p>
<p>I don’t think I follow your logic. Can you explain this further?</p>
<p>You know how the axis of sym equation is -b/2a well since I was already given the points 1,1. Now I can use the -b =1 and dividing by 2a can be ignored and solving for b gives -1.</p>
<p>man, I dont even know why I came up with that. I was just frustrated trying to get an answer haha. :P</p>