find tangent and perpendicular equation at the x axis for...

Yes Stu, if two lines are perpendicular then the product of their gradients will be -1

$$m_1m_2=-1\\\\ therefore\\\\ m_2=\dfrac{-1}{m_1}$$

 

The normal to a curve is the name given to a line that is perpendicular to the tangent of the curve at that point.

That is;  The normal is perpendicular to the tangent.

Hi Stu,
(a)
the graph y=e ^x has a very special property. The y value is equal to the gradient of the tangent.
so
dy/dx also equals e ^x
Reiterating:
If y=e ^x then y'=e ^x

This is your question
f(x)=1-e ^x
You are asked to find the gradient of the tangent to the curve at the point where it crosses the x axis.
In this case y has been replaced with f(x). [Call it f(x) but treat it the same as if it was y]
Any curve crosses the x axis when y=0 or in this case, f(x)=0 This is the x intercept.
Find the x intercept
0=1-e ^x
e ^x=1
x=0
now
f(x)=1-e ^x
f'(x)=0-e ^x = -e ^x
when x=0
f'(0)=-e ⁰ = -1

this is the end of part a. I haven't done b or c yet.
Part b is missing the point - do you know what it is supposed to be?

...
Part b is missing the point - do you know what it is supposed to be?[/quote]


I think it means the point at which f(x) crosses the x axis (i.e. the point (0, 0)).

Part (b)
We want to find a tangent line to f(x) = 1 - e^x where it crosses the x axis. As posted earlier, this point is (0,0)

So, the derivative of this unction gives the slope (gradient) at any point on the curve. So, the derivative is given by f'(x) = -e^x. So, the slope of the tangent line at this point is just -(e^0) = -1. So, using the point-slope form we have:

y - 0 = (-1)(x - 0)

y = -x

So, in Part (c), we want to find the normal line to this. Then the slope of the normal line is just 1. And using the point-slope form in a similar manner we have:

y - 0 = (1)(x - 0)

y = x