When a is positive, the parabola opens up and has a minimum When a is negative, the parabola opens down and has a maximum
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2 KEY CONCEPTS For a quadratic relation of the form y = ax 2 + c, the maximum or minimum value occurs at c, which is the y-intercept. When a is positive, the parabola opens up and has a minimum When a is negative, the parabola opens down and has a maximum
3 EXAMPLE 1 Finding the Maximum or Minimum For each relation, identify the following: The a value The direction of the opening of the parabola Maximum or minimum and its value a = 5 a = 1 a = 3 a is positive Parabola opens up a is positive Parabola opens up a is negative Parabola opens down Parabola opens up Minimum value of 0 Parabola opens up Minimum value of 16 Parabola opens down Maximum value of 27
4 EXAMPLE 2 Finding the Zeros of a Quadratic Relation Without graphing, find the zeros of the following relations. Be sure to factor first!!! Recall: When finding zeros, let y = 0 and solve for x (a) y = x = x = ( x + 5 )( x 5 ) x 2 x 25 5 Factor by difference of squares *Write down two sets of brackets * Place a + in the first bracket and in the second bracket. *Take the square root of the first term. Place this answer in the first term of each bracket. *Take the square root of the second term. Place this answer in the second term of each bracket.
5 EXAMPLE 2 Finding the Zeros of a Quadratic Relation Without graphing, find the zeros of the following relations. Be sure to factor first!!! Recall: When finding zeros, let y = 0 and solve for x (a) y = x = x = ( x + 5 )( x 5 ) 1 st zero 0 = x = x 2 nd zero 0 = x 5 5 = x Factor by difference of squares *Write down two sets of brackets * Place a + in the first bracket and in the second bracket. *Take the square root of the first term. Place this answer in the first term of each bracket. *Take the square root of the second term. Place this answer in the second term of each bracket.
6 EXAMPLE 2 Finding the Zeros of a Quadratic Relation Without graphing, find the zeros of the following relations. Be sure to factor first!!! Recall: When finding zeros, let y = 0 and solve for x (b) y = x = x = ( x + 4 )( x 4 ) x 2 x 16 4 Factor by difference of squares *Write down two sets of brackets * Place a + in the first bracket and in the second bracket. *Take the square root of the first term. Place this answer in the first term of each bracket. *Take the square root of the second term. Place this answer in the second term of each bracket.
7 EXAMPLE 2 Finding the Zeros of a Quadratic Relation Without graphing, find the zeros of the following relations. Be sure to factor first!!! Recall: When finding zeros, let y = 0 and solve for x (b) y = x = x = ( x + 4 )( x 4 ) 1 st zero 0 = x = x 2 nd zero 0 = x 4 4 = x Factor by difference of squares *Write down two sets of brackets * Place a + in the first bracket and in the second bracket. *Take the square root of the first term. Place this answer in the first term of each bracket. *Take the square root of the second term. Place this answer in the second term of each bracket.
8 EXAMPLE 3 Finding the Equation of a Quadratic Relation The graph of the quadratic relation y = ax passes through the point (2, 21). Find the value of a and the equation for this quadratic relation. x y Substitute the point (2, 21) into the relation for x and y y ax a(2) a a 16 4a a For the equation Substitute a = 4 into the relation y y ax 2 5 4x 2 5
9 EXAMPLE 4 Application: Hitting a Golf Ball Cheetah hits a golf ball from the top of a mountain. The path of the ball can be represented by the relation h = 5t , where h represents the height of the ball in metres and t represents the time, in seconds. (a) From what height was the ball hit from? Method 1: Look at the y-intercept (or in this case, the h-intercept) h = 245 The ball was hit from 245 metres above the ground. Method 2: Solve for the height of the ball before it was hit Let t = 0 and solve for h h = 5t = 5(0) = = 245
10 EXAMPLE 4 Application: Dropping a Ball Cheetah hits a golf ball from the top of a mountain. The path of the ball can be represented by the relation h = 5t , where h represents the height of the ball in metres and t represents the time, in seconds. (b) How many seconds was the ball in the air before hitting the ground? When the ball hits the ground, h = 0 h 5t t t t 49 t 7 t 2 2 Solve for t The ball was in the air for 7 seconds before hitting the ground. Take the square root of both sides
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