Finding Equations of Curves (1 Viewer)

[flon]

hey guys help us out with this question

find the equation of the curve that is always concave upwards with a stationary point at (-1,-2) and y intercept 3.

thanks

 

[independantz]

given: f'(1)=0, f(1)=-2, satisfies (0,3)

consider: f(x)=ax^2+bx+c

2a+b=0 using f'(1)=0
a+b+c=-2, using f(1)=-2
c=3, since the curve passes through (0,3)

solving simultaneously...

a=5,b=-10,c=3 these may be wrong as i did it in my head...

therfore eq of curve : 5x^2+10x+3

 

[ianc]

hi there

always concave upwards = smiley faced parabola

this means it is in the form of y=ax^2 + bx + c

note that because it is concave up, this means that a must be postitive.

so it's just a matter of finding a,b,c


we are given a stationary point. this is really helpful!

• we know that the parabola must pass through the stationary point. so substitute in x=-1 and y=-2 into our equation
  
  
• differentiate our equation to get dy/dx = 2ax + b
  the stationary point is when dy/dx=0, so substitute in x=-1 into 2ax+b=0
  
  
• y intercept is 3: this just gives us another point the curve passes through. so substitute in x=0, y=3 into our equation.
  

so by now you should have 3 equations, with 3 unknowns (a,b,c) -- so just solve them and you've worked it out!