# Math Help - tangent equation

1. ## tangent equation

hi, my question is: find and equation of the tangent to the following curve at the point where x = 0

$y = 3x^4 + 5x - 2$

first i differentiated so i could find the gradient: $\frac{dy}{dx} = 12x^3 + 5$ then subbing x = 0 into the equation you get 5 for the gradient.
using x = 0 you can also see that from the original equation y = -2

then using the equation formula $y - y1 = m(x - x1)$ i got $y + 2 = 5(x - 0) \implies y = - 2$ so the equation looks to me like it should be y = -2

the answer in the back gives $y = 5x - 2$ for the equation

can someone show me what i'm doing wrong please,
thankyou

ps what do you have to type into latex to get the 1's in $y - y1 = m(x - x1)$ to appear small and at the bottom of the x and y if you know what i mean? like the same size as "to the power of" but at the bottom

2. Originally Posted by mark
hi, my question is: find and equation of the tangent to the following curve at the point where x = 0

$y = 3x^4 + 5x - 2$

first i differentiated so i could find the gradient: $\frac{dy}{dx} = 12x^3 + 5$ then subbing x = 0 into the equation you get 5 for the gradient.
using x = 0 you can also see that from the original equation y = -2

then using the equation formula $y - y1 = m(x - x1)$ i got $y + 2 = 5(x - 0) \implies y = - 2$ so the equation looks to me like it should be y = -2

the answer in the back gives $y = 5x - 2$ for the equation

can someone show me what i'm doing wrong please,
thankyou

ps what do you have to type into latex to get the 1's in $y - y1 = m(x - x1)$ to appear small and at the bottom of the x and y if you know what i mean? like the same size as "to the power of" but at the bottom
How does $y+2 = 5(x-0)$ imply $y = -2$...? Check your algebra!

To make it appear as you want, use the underscore symbol. For example: x_1, y_1, z_1 yields: $x_1, y_1, z_1$. If you want more than one character there, wrap them in {} symbols, ie: z_{abc} gives: $z_{abc}$.

3. yeah i can't believe i made that mistake, how stupid. thanks for the help