Stationary Points and finding variables?

[Dramacon]

Stationary Points and finding variables? :)

Homework Statement

The curve with the equation y=ax^2 + bx+c has a stationary point at (1,2). When x=0, the slope of the curve is 45 degrees. Find a, b, c.

Homework Equations

I'm not sure how to create an equation that involves the 45 degrees

The Attempt at a Solution

There are three equations to solve for the three variables, but so far I only have 2 equations:

2= a+b+c
and
2a+b=2

DW, SOLVED IT! :)

 

[Simon Bridge]

Well done!

Sometimes explaining the problem to someone else will lead to a solution :)
Want to share how you did it?

 

[Dramacon]

the derivative of a function is its gradient, which is also defined as rise/run. In trig, rise/run = tanθ!

therefore tan45°=1

And then just use one of the aforementioned methods of substitution into y'.

:)

 

[HallsofIvy]

Yes. The fact that the function has a stationary point at (1, 2) tells you two things:
First that the value of the function at x= 1 is 2: $a(1)^2+ b(1)+ c= a+ b+ c= 2$.
Second that the derivative there is 0: 2a(1)+ b= 2a+ b= 0.

The fact that the tangent line, at x= 0, is at 45 degrees to the x-axis tells you that the slope of the tangent line there is tan(45)= 1: 2a(0)+ b= 1.

Your three equations are b= 1, 2a+ b= 0 so 2a+ 1= 0, and a+ b+ c= 2 so a+ c+ 1= 2.

 

[Simon Bridge]

Outstanding Dramacon: I figured I'd give you the glory of explaining it :) There is a chance someone else will google to your problem when they are looking for a solution to theirs. Now you've just helped them.

HallsofIvy, is a veteran, and has included the complete solution.

I prefer the "rise over run" version since it uses what "slope" means in relation to the gradient ... though the trig is a useful addition so this is totally not a criticism. The tangent of an angle is also rise over run - of course.