Calculus question kinda confusing (1 Viewer)

[MartinsPlace]

Find the x-coordinate of the point on the curve y = 3x² - 2x - 4 where the tangent is perpendicular to the line 2x - 3y + 7 = 0.

Can someone tell me where I did a mistake or if I did the whole thing completely wrong.

(dy/dx) = 6x - 2

3y = 2x + 7
y = (2/3)x + (7/3)

(6x - 2)(-3/2) = - 1
[(- 18x + 6)/(2)] = - 1
- 18x + 6 = - 2
- 18x = - 8
x = 4/9

I know this is wrong, but what did I do wrong?

 

[Shadowdude]

could you please explain all your steps and what you were trying to achieve

because you just skip explanation and i'm finding it difficult to figure out what you're trying to do

 

[Carrotsticks]

If you wanted to do the (6x-2)(XXXXX) = -1 approach, then the (XXXX) would have to be 2/3, not -3/2.

By making it -3/2, you already made it perpendicular, so you're just finding the perpendicular of the perpendicular, which is the tangent.

 

[MartinsPlace]

Okay

1) Find the first derivative of y = 3x² - 2x - 4, which is y' = 6x - 2
2) Re-arrange the line 2x - 3y + 7 = 0 into y = mx + b so I can find the gradient 'm' which is 2/3
3) Since for perpendicular m1.m2 = - 1, thus (6x - 2)x(2/3) = - 1
4) Multiplying them, (12x - 4)/(3) = - 1
5) 12x - 4 = - 3
6) 12x = 1
7) x = 1/12

Sorry, I just realised I accidentally wrote the gradient from a previous question during my working out & completely stumbled me.

Thanks anyways!!

 

[MartinsPlace]

If you wanted to do the (6x-2)(XXXXX) = -1 approach, then the (XXXX) would have to be 2/3, not -3/2.

By making it -3/2, you already made it perpendicular, so you're just finding the perpendicular of the perpendicular, which is the tangent.

Yup i realised just now what i did. Thank you carrotstick!