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Mathematics Dictionary
Dr. K. G. Shih
Vertex of Quadratic Function
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Q01. Vertex of y = a*x^2 + b*x + c
Defintion and properties
  • Vertex of y = a*x^2 + b*x + c has maximum or minimum at vertex.
  • At vertex of y = a*x^2 + b*x + c the slope is zero.
  • The vertex is a minimum if a is negative.
  • The vertex is maximum if a i negative.
Formula
  • Vertex is at (xv,yv).
  • xv = -b/(2*a).
  • yv = F(xv) = -(b^2-4*a*c)/(4*a)
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Q02. How to find the vertex of y = F(x) = a*x^2 + b*x + c ?
Method 1 : Using the completing square
  • y = a*(x^2 + b*x/a + c/a)
  • y = a*(x^2 + b*x/a + (b/(2*a))^2 - (b/(2*a))^2 + c/a)
  • y = a*(x+b/(2*a))^2 - a*((b/(2*a))^2 - c/a)
  • y = a*(x+b/(2*a))^2 - b^2/(4*a) - c
  • y = a*(x+b/(2*a))^2 - (b^2-4*a*c)/(4*a)
  • If y is minimum or maximum, then (x+b/(2*a)) = 0.
  • Hence xv = -b/(2*a) and yv = -(b^2-4*a*c)/(4*a).
Method 2 : Using Derivative = 0
  • The derivative of y = a*x^2 + b*x + c is y' = 2*a*x + b.
  • Since y' = 0 at vertex, hence xv = 2*a*xv + b = 0
  • Hence xv = -b/(2*a) and yv = F(xv)
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Q03. Find Vertex of y = x^2 - 6*x + 8 in which quadrant

  • Vertex of y = x^2 - 6*x + 8 is at (xv,yv).
  • xv = -b/(2*a) = -(-6)/(21) = 3.
  • yv = -(b^2 - 4*a*c)/(4*a) = -(36 - 4*1*8)/(4*1) = -1.
  • The vertex is in 4th qudrant.
  • Note : We can also use yv = F(xv) = (3)^2 - 6*3 + 8 = -1.
Example : Find Vertex of y = x^2 + 6*x + 8 in which quadrant
  • Vertex of y = x^2 + 6*x + 8 is at (xv,yv).
  • xv = -b/(2*a) = -(6)/(21) = -3.
  • yv = -(b^2 - 4*a*c)/(4*a) = -(36 - 4*1*8)/(4*1) = -1.
  • The vertex is in 3rd qudrant.
  • Note : We can also use yv = F(xv) = (-3)^2 - 6*(-3) + 8 = -1.
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Q04. Find the Focus of the paranola y = a*x^2 +b*x + c
Formula to find focus of parabola
 Example : Find focus of y = x^2 - 6*x + 8
  • Find vertex.
    • xv = -b/(2*a) = -(-6)/(2*1) = 3.
    • yv = F(3) = 3^2 - 6*3 + 8 = -1.
  • Find distance from focus to directrix.
    • D = 1/(2*a) = 1/2.
    • Hence equation of directrix is y = yv - D/2
  • Find focus.
    • xf = xv = 3.
    • yf = yv + D/2 = -1 + (1/2)/2 = -3/4
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Q05. What is the directrix of a parabola y = a*x^2 + b*x + c ?
  • The distance of directrix to focus is D = 1/(2*a).
  • The directrix is perpendiular to the principal axis of the parabola
  • The equation of the directrix is y = yv - D/2 if (xv,yv) is the vertex.
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Q06. Example : y = x^2 -6*x + 8
Find vertex
  • xv = -b/(2*a) = -(-6)/(2*1) = 3.
  • yv = F(3) = 3^2 - 6*3 + 8 = -1.
Find equation of principal axis
  • Equation of principal axis is x = xv = 3.
Find equation of directrix
  • Distance fro focus to directrix is D = 1/(2*a) = 1/2.
  • Equation of directrix is y = yv - D/2 = -1 + (1/2)/2 = -1.25.
Find focus
  • xf = xv = 3
  • yf = yv + D/2 = -1 + (1/2)/2 = -0.75
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Q07. Reference
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Q08. Answer
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Q09. Answer
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Q10. Answer
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