Standard Form

By: Harman Virk

Learning Goals

1) I am able to complete the square


Example: Rewrite y= x^2 + 8x + 5 in the form y = a (x - h)^2 + k


y = (x^2 + 8x) + 5 <-- put a bracket around the first two terms and take the common out


8/2 = 4 4^2=16 <-- divide 'B' by 2 and square it


y = (x^2 + 8x +16 -16) + 5 <-- Put the squared number and add then subtract. There should be a perfect square in the brackets.


y = (x + 4)^2 + 5 -16 <-- The bracket has become a perfect square


y = (x + 4) ^2 - 11 <-- You have completed the square




2) Quadratic Relations of the form: y = a(x - r) (x - s)


Example: Describe the graph of the quadratic relation y = 2(x + 1) (x - 7)


x - int --> -1 and 7 <-- These are the y-intercepts of the vertex


AOS = (-1 + 7) / 2 = 3 <-- This is the x - value in the vertex


y = 2(3 +1) (3 - 7) = -32 <-- This is how you find the y - value of the vertex


(3, -32) <-- This is the vertex


y = 2(0 + 1) (0 - 7)

y = 2 (1) (-7)

y = -14 <-- This is the y-intercept

Summary of the Unit

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Word Problem

The path of a ball is modeled by the equation y = -x^2 +2x + 3, where x is the horizontal distance, in metres, from a fence and y is the height, metres, above the ground.


a) What is the maximum height of the ball, and at what horizontal distance does it occur?


y = -x^2 + 2x + 3

y = -(x - 2x) + 3


-2/2 = -1^2= 1

y = -(x^2 - 2x + 1 - 1) + 3

y = -(x - 1)^2 + 4


Therefore, the maximum height is 4m and it occurs at 1 second.