Given f(x)=3x^2+2x-1 , answer the following questions: 1.What is the EQUATION of the line that is tangent to the graph at ? 2.At what point (x, y) on the graph is its slope of 4?
Do you know how to take the derivative?
yes, I do
Actually, I did first question.
But I do not know how to do the second one.
Gimme a sec, I'll brb and explain it to you.
So you want to take the derivative as you normally do. What does that give you?
I just want to make sure you got the right answer for part one, because you need it for the second part
f'(X)= 6x+2
Yep, and that equation gives you the slope at any given x on the original line. Does that give you any idea on how to find the coordinates when the slope is 4?
set f'(x) to zero ??????
and solve for x???
That's only if you want to find the max or min of the line. Maxes and mins have a slope of 0. So this time what would you need to do?
find f'(4)????
No, that would be setting x equal to 4. Give it some thought, When you find maxes and mins, you solve for x by setting the derivative=0. What do you want to do instead this time?
substitute x with 4????
I do not know:-(
Don't just make guesses, make comparisons. When you solved for maxes and mins, what did you do?
solve for x by setting the derivative=0
Yes, and what is the slope of the line at the max or min?
0
So this time what slope are we concerned with?
4
Now you tell me what needs to be done. You have all the information in front of you.
This is not a trick question, just put two and two together
(1/3;4)???????
is it correct?
Okay it looks like you got the first half right. But after you set the derivative=4 and solve for x, we want to find the position on the original line where that slope of 4 is. You want to plug x back into the original and solve for y.
should be (1/3; 0)
Right???
What did you get for x when you set the derivative=4?
1/3
Oh nvm I made the mistake. Yes it is (1/3, 0). See, it wasn't that tough right? Just gotta make the connections
it was easy after you gave me a hint. I really appreciate your help. Thank you again.
Yep, glad you understand a little bit more about derivatives now (hopefully)
yep, it is more clearly now.
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