Oh so you're looking for the y-axis intercept then. I don't understand why your boss would give you a question you cannot even start to answer
Anyway, for a function y=ax
n where a and n are constants, the derivative is y'=anx
n-1
You will have to use the point-gradient formula for a line, which is
y-y
0=m(x-x
0) where (x
0,y
0) is a point on the line and m is the gradient of the line.
For one line we know the x and y intercept because it's what is required of us, mainly (0,150) and we don't know the gradient, so we will leave it as m.
For the next equation we will use the other info we can determine which is the point of contact between the line and the curve, we don't know this point but since it lies on the curve we can say it is (x
1,ax
1n) where x
1 is the x-coordinate we are looking for.
Now from the derivative equation, it is telling us what the gradient of the tangent is at an x-coordinate on the curve, so we will substitute x
1 for x since it is the same point, so we have the gradient of the curve is anx
1n-1
Setting up our second line equation, we have
y-ax
1n=anx
1n-1(x-x
1)
So you have two line equations, and you have a whole bunch of constants and stuff that you can throw in later (I prefer to keep any large ugly numbers out of equations so it's easier to see). You need to find out what x
1 is because then you can find out the point of intercept between the line and curve, and then the gradient, which is finally what you need. Do you know how to solve equations simultaneously?
