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the tangent line occurs at the vertex of the this particular function which happens to be a parabola
Okay so first I would have to derive my equation right?
So I would get 9x-3 and then what would I do?
Well from there...you want a tangent line that is horizontal...and since a tangent line is the slope at any given point...that would mean the slope would be 0 right? So if we have the derivative of the function...we need to find where the derivative = 0
So it would be -3 at 0
No I mean we have the derivative...and we want the derivative to equal 0 \[\large 9x - 3 = 0\] solve for 'x' :)
Oh my bad to it would be 3/9.
Mmhmm...or 1/3 right :P So that means that at x = 1/3 ...we will have a horizontal tangent line
And if I wanted to find the equation for the tangent line at 2 would i just plug in 2 into my derived equation?
But we probably should figure out where that will be in terms of 'y' now... So plug in your x = 1/3 into your original equation and solve for 'y'
I got 1.5
Good...so just to finish up the first question At (1/3 , 1.5) there will be a horizontal tangent line
And as far as your second question...yes you would just plug in x = 2 and solve for y in your derived ezpression
Cant spell apparently >.< "expression"
Okay thank you so much!
No problem! :)