You would best do this by completing the square on the quadratic and subbing in the vertex they give you, like this:

Since the y coordinate of the vertex is 1/2 and we need to solve for c, set the

equal to 1/2 and solve for c, which is the whole point of what we are doing, right?

and


and c = 4. So that's our answer!
Answer:
f'(1)=150ln(1.5)
Step-by-step explanation:
I'm not sure why you would need a table since the limit definition of a derivative (from what I'm remembering) gives you the exact formula anyway... so hopefully this at least helps point you in the right direction.
My work is in the attachment but I do want to address the elephant on the blackboard real quick.
You'll see that I got to the point where I isolated the h's and just stated the limit equaled the natural log of something out of nowhere. This is because, as far as I know, the way to show that is true is through the use of limits going to infinity. And I'm assuming that you haven't even begun to talk about infinite limits yet, so I'm gonna ask you to just trust that that is true. (Also the proof is a little long and could be a question on it's own tbh. There are actually other methods to take this derivative but they involve knowing other derivatives and that kinda spoils a question of this caliber.)
Answer:
216
Step-by-step explanation:
Answer: x = -0.3
Step-by-step explanation:
<u>Given expression</u>
1 - 3 (9x + 3) = 3x + 1
<u>Expand parentheses and apply the distributive property</u>
1 - 27x - 9 = 3x + 1
<u>Combine like terms</u>
(1 - 9) - 27x = 3x + 1
-8 - 27x = 3x + 1
<u>Subtract 3x on both sides</u>
-8 - 27x - 3x = 3x + 1 - 3x
-8 - 30x = 1
<u>Add 8 on both sides</u>
-8 - 30x + 8 = 1 + 8
-30x = 9
<u>Divide - 30 on both sides</u>
(-30)x / (-30) = 9 / (-30)
(-30)x / (-30) = 3 / (-10)

Hope this helps!! :)
Please let me know if you have any questions