# Math Help - At what point on the given curve is the tangent line parallel to the line 3x - y = 3?

1. ## At what point on the given curve is the tangent line parallel to the line 3x - y = 3?

At what point on the given curve is the tangent line parallel to the line 3x - y = 3?

y = 2 + 2ex3x

get x and y.

what I did so far:

rearranged 3x=y=3 to get y=3x-3, which shows the slope is 3, so i need to find where my tangent line has a slope of 3 on the line 2+2e^x-3 i think.
How would i proceed? Thanks.

2. Originally Posted by Evan.Kimia
At what point on the given curve is the tangent line parallel to the line 3x - y = 3?

y = 2 + 2ex3x

get x and y.

what I did so far:

rearranged 3x=y=3 to get y=3x-3, which shows the slope is 3, so i need to find where my tangent line has a slope of 3 on the line 2+2e^x-3 i think.
How would i proceed? Thanks.
It is commonly known that if everything is permissable (defined, differentiable, etc.) the slope of the tangent line at $x=x_0$ for a function $f(x)$ is just $f'(x_0)$

So you want to find the point $x_0$ such that $\frac{d}{dx}\left[2+2e^x-3x\right]=3$. Can you do that?

3. ok, so would i do this....

find deriv. of 2+2e^x-3x

= 0 + 2e^x (product rule) -3 = 3

then add 3 both sides, get 2e^x = 6

divide by 2, get e^x=3

ln(3)=1.0986?

4. Originally Posted by Evan.Kimia
ok, so would i do this....

find deriv. of 2+2e^x-3x

= 0 + 2e^x (product rule) -3 = 3

then add 3 both sides, get 2e^x = 6

divide by 2, get e^x=3

ln(3)=1.0986?
Exactly! One minor note....how did you get $\frac{d}{dx}\left[e^x\right]=e^x$ via product rule?

5. I got it. For those who were as confused as i was, just plug in the x value into the equation of the curve to get your y value.

6. my bad, lazy typing. I mean product rule at first for 2e^x, then i know the derivative of e^x is itself, which is then multiplied by 2 according to the constant multiple rule.