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Quadratic Functions


17 EXAMPLE 12


Find where the function with equation y = –2x 2 + 5x + 3, x ∈ , crosses the axes. Find the co-ordinates of its vertex and the equation of its axis of symmetry.


Solution y = –2x 2 + 5x + 3, x ∈ 


a = −2, b = 5, c = 3 Crosses y-axis: x = 0 y = 3


It crosses the y-axis at (0, 3). Crosses x-axis: y = 0 2x 2 – 5x – 3 = 0 (2x + 1)(x – 3) = 0


x = − 1 __


2 , 3


It crosses the x-axis at 3 +


Vertex V: x =


or x = − b


x = 5 __


∴ V


( 5 __


________ 2


( − 1 __


___ 2a = − 5


4 : y = −2


4 , 49 ___


8 )


Equation of the axis of symmetry: x = 5


__ 4


or 4x − 5 = 0 EXERCISE 6


1. Draw the following quadratic functions on graph paper in the given domain for x ∈ :


Note: This is also Activity 4. The activity supplies you with the appropriate grids.


(a) f (x) = 2x 2 − 5x + 1, −1 ≤ x ≤ 3 (b) f (x) = − 2x 2 + 3x + 4, −2·5 ≤ x ≤ 3 (c) f (x) = x 2 − 2x + 1, −1 ≤ x ≤ 3 (d) f (x) = −4x 2 + 4x − 1, −1 ≤ x ≤ 1 (e) f (x) = 2x 2 + 3x + 2, −2·5 ≤ x ≤ 1 (f) f (x) = − x 2 + 2x − 3, −1 ≤ x ≤ 3


2. Find the points at which the following quadratic functions cross the axes. Give all answers as rational or irrational numbers. If the curve does not cross the x-axis, say so.


(a) y = x 2 – 1 (b) y = 3x 2 + 1 (c) y = x 2 – 5x + 6 (d) y = (2 – x)(x – 1) (e) y = 2(2x – 1)2


(f) y = –3x 2 + 8x – 4 (g) y = 4 – (x – 1)2 (h) y = 4(x – 2)2 – 9 (i) y = x 2 – 5x + 7 (j) y = 2x 2 + 7x – 11


3. For all the functions in Question 2, fi nd the co-ordinates of the maximum or minimum value of the quadratic function and the equation of the axis of symmetry.


4. By calculating b 2 – 4ac, state for each quadratic function, y = ax 2 + bx + c, whether the function crosses the x-axis at one point or at two different points:


(a) y = x 2 – 5 (b) y = x 2 – 7x + 12 (c) y = 15 + 14x – 8x 2 (d) y = 4(3x – 2)2 (e) y = 4x 2 – 1 (f) y = 4x 2 – 7x – 1 (g) y = (2x – 1)(3x + 5) (h) y = 3(2 – 3 x) 2


( 5 __


2 )


_____ 2(−2) = 5


4 ) + 5 2


__ 4


( 5 __


4


) + 3 = 49 ___


8


( − 1 __


2 , 0 ) and (3, 0).


= 5 __


4 [Using the midpoint formula]


261


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