finding a parallel line.

[Synergy]

Hey,

I have a question that is asking for the point at which a tangent on a parabola will be parallel to a line with the equation: y = 14x -16

EDIT: (Sorry but my information was wrong first time)

The equation of parabola is: y = 2x^2 +3x

now it ask me for the point on the parabola at which the tangent is parallel to the line y = 14x - 6!

***From another part in same question I found the slope of some tangent to be 4a+3, not sure if I should or could use that in this question!!!***

[Synergy]

Can someone at least tell me how I would formulate an equation of the tangent using the slope 4a+3, where x is a?

[Synergy]

Ok I got a bit ahead now, what I did was this:

I know that for tangent to be parallel to line y = 14x + 6 their slops need to be equal! So I know thats lope of tangent is 4a+3, so whats a that will make 4a+3 = 14.

14 = 4a+3
14 - 3 = 4a
a = 11/4 <=> a is also our x
so x = 11/4 which is our first point!

So now I know that the point is (11/4, Y). I need to find Y somehow but not sure how!!! Can someone push me in the right direction???

[moeur]

Just plug your x back into the parabola equation to get Y
Y = 2(11/4)^2 + 3(11/4)

[Synergy]

Ok, I somehow got Y but the way I got it might not be right! I did this:
equation of parabola = equation of tangent

y = E.O.T

so 2x^2 + 3x = 11/4x +b

2(11/4)^2 + 3(11/4) = 11/4(11/4) + b
308 / 8 =121 / 8 + b
so 308 - 121/8 = b

which is 187/8 = b
at the back index the answer is that! So I got correct y but I say the way I got it is incorrect since I took slope of tangent to be 11/4 when I think it should be 14! correct or is the way I did it right???

[Synergy]

Just plug your x back into the parabola equation to get Y
Y = 2(11/4)^2 + 3(11/4)

But the you get 308/8, while at the back they have 187/8!!!

[Synergy]

Ok I think they got wrong answer at the back, So yeah I just did as you said plugging x into equation of parabola1 Thanks for the help guys!!!