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.
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
i dont really get this equation anyone know were to start
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
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.
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
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!
In general, let be a differentiable function.
Then at the point tangent line of is for all .
And let for some with be a line, which is supposed to be perpendicular to the line .
Then, we have to determinate in such a way that is perpendicular to ; that is, .
If as inverse function (as in this case), then
and so
.
The rest is just simple calculation.
I hope you will find this clearifying.