# Thread: Find the equation of the tangent line(s)

1. ## Find the equation of the tangent line(s)

Find the equation of the tangent line(s),which is perpendicular to the line x+6y+5=0, to the graph of F(x) at the tangent point(s) (x,y)
Hint:Find the slope first then use y-y=m(x-x)

The equation is $\displaystyle f(x)= 2x^3-3$

i dont really get this equation anyone know were to start

2. Find the slope of x+6y+5=0
Find the derivative of f(x)

what is the relationship of the slope between 2 lines that are perpendicular?

Set that slope equal to f'(x)

Hope this helps. Ask if you are still confused.

3. Find the slope of x+6y+5=0

$\displaystyle m=-1/6$

Find the derivative of f(x)

$\displaystyle 6x^2$

what is the relationship of the slope between 2 lines that are perpendicular?

I believe the slopes the same

Set that slope equal to f'(x)

$\displaystyle -1/6=6x^2$

Still confused

4. Originally Posted by john doe
Find the slope of x+6y+5=0

$\displaystyle m=-1/6$

Find the derivative of f(x)

$\displaystyle 6x^2$

what is the relationship of the slope between 2 lines that are perpendicular?

I believe the slopes the same

Set that slope equal to f'(x)

$\displaystyle -1/6=6x^2$
Slopes are the same when they are parallel.
When 2 lines are perpendicular, their slopes are opposite reciprocal of each other.

So you would set $\displaystyle 6x^2=6$

Find x and put it back into the original equation to find y and that will be your point.

Now just use the point-slope equation since you know the point and the slope.

5. hows this look

$\displaystyle y+1=6(x-1)$

6. Originally Posted by john doe
hows this look

$\displaystyle y+1=6(x-1)$
There should be 2 lines.

7. i dont understand

8. Originally Posted by john doe
i dont understand
haha sorry it's late here, I forgot to add in the explanation.

$\displaystyle 6x^2=6$

x=1, -1

9. i must be losing it because i dont know what that means

my graph kinda turned out with a parabola and a tangent line
going though at 1,-1

its kinda 3 aclock

10. Let's back up a bit.

Here is your question: Find the equation of the tangent line(s),which is perpendicular to the line x+6y+5=0, to the graph of F(x) at the tangent point(s) (x,y)

so we need to find at where does F(x) have slopes of 6.

So we did $\displaystyle F'(x)=6x^2=6$
the solution would be x=1 and -1 because x is squared

That means the function F(x) has 2 points where the slope is 6 that is at (1, F(1)) and (-1, (F(-1)).

This means there are 2 lines which satisfies the question.

Hope this clears things up!

11. I think i see it at
so that would look like a x^3 graph shifted to the bottom
theres a slope 6 that hits the graph 2 times 1 and -1

12. Originally Posted by john doe
I think i see it at
so that would look like a x^3 graph shifted to the bottom
theres a slope 6 that hits the graph 2 times 1 and -1
Hmm, I'm not quite sure what you mean.

The slope of x+6y+5=0 would be -1/6

Perpendicular slope=6

$\displaystyle f'(x)=6x^2=6$
$\displaystyle x= 1,-1$

$\displaystyle f(1)=-1$
$\displaystyle f(-1)=-5$

so the 2 lines would be

$\displaystyle y+5=6(x+1)$
$\displaystyle y+1=6(x-1)$

All I was trying to tell you after your first solution is that there is more than one answer.

13. alright thax for the help

i dont fully understand it but

i appreciate all the time your talking

this late at night to explain to me

(ill take a look at it tomorrow)

14. In general, let $\displaystyle f:[t_{0},\infty)\to\mathbb{R}$ be a differentiable function.
Then at the point $\displaystyle s\in(t_{0},\infty)$ tangent line of $\displaystyle f$ is $\displaystyle g(t):=f^{\prime}(s)(t-s)+f(s)$ for all $\displaystyle t\in[t_{0},\infty)$.
And let $\displaystyle h(t):=at+b$ for some $\displaystyle a,b\in\mathbb{R}$ with $\displaystyle a\neq0$ be a line, which is supposed to be perpendicular to the line $\displaystyle g$.
Then, we have to determinate $\displaystyle s$ in such a way that $\displaystyle g$ is perpendicular to $\displaystyle h$; that is, $\displaystyle a*f^{\prime}(s)=-1$.
If $\displaystyle f^{\prime}$ as inverse function $\displaystyle (f^{\prime})^{-1}$ (as in this case), then
$\displaystyle s=(f^{\prime})^{-1}(-1/a)$
and so
$\displaystyle g(t)=-\frac{1}{a}\big(t-(f^{\prime})^{-1}(-1/a)\big)+f\big((f^{\prime})^{-1}(-1/a)\big)$.
The rest is just simple calculation.
I hope you will find this clearifying.