Sketching quadratic functions
To sketch a quadratic function we need to identify where possible:
The y intercept (0, c)
The roots by solving ax2 + bx + c = 0
The axis of symmetry (mid way between the roots)
The coordinates of the turning point.
The shape: 2 ax bx c
If 0 then a If 0 then a
21. Sketch the graph of 5 4y x x
The shape
The coefficient of x2 is -1 so the shape is
The Y intercept
(0 , 5)
The roots25 4 0x x
(5 )(1 ) 0 x x
(-5 , 0) (1 , 0)
The axis of symmetry
Mid way between -5 and 1 is -2
x = -2
The coordinates of the turning point
When 2, 9x y (-2 , 9)
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
(-2 , 9)
Completing the squareThe coordinates of the turning point of a quadratic can also be found by completing the square.
This is particularly useful for parabolas that do not cut the x – axis.
REMEMBER
2 2 can be written in the form ( ) +q. y ax bx c y a x p
Axis of symmetry x pCoordinates of turning point ( , ).p q
2
1. Find the equation of the axis of symmetry and the coordinates of the
turning point of 2 8 9.y x x 22 8 9y x x
22( 4 ) 9x x 22( 2) 9 8x 22( 2) 1x
Axis of symmetry is x = 2
Coordinates of the minimum turning point is (2 , 1)
2
2. Find the equation of the axis of symmetry and the coordinates of the
turning point of 7 6 .y x x 2 6 7y x x 2( 6 ) 7x x
2( 3) 7 9x 2( 3) 16x
Axis of symmetry is x = 3
Coordinates of the maximum turning point is (3 , 16)
Solving quadratic equations
Quadratic equations may be solved by:
The Graph
Factorising
Completing the square
Using the quadratic formula
21. Solve 6 15 0x x (2 3)(3 5) 0x x
2 3 0x 3
2x
3 5 0x 5
3x
22. Solve 4 1 0x x
This does not factorise.
2 0ax bx c 2 4
2
b b acx
a
1, 4, 1a b c
4 16 4
2x
4 20
2
4 2 5
2
2(2 5)
2
2 5 or 2 5
Quadratic inequations
A quadratic inequation can be solved by using a sketch of the quadratic function.
21. For what values of is 12 5 2 0?x x x
First do a quick sketch of the graph of the function.
212 5 2 (4 )(3 2 )x x x x Roots are -4 and 1.5
Shape is a
-4 1.5 The function is positive when it is above the x axis.
34
2x
22. For what values of is 12 5 2 0?x x x
First do a quick sketch of the graph of the function.
212 5 2 (4 )(3 2 )x x x x Roots are -4 and 1.5
Shape is a
-4 1.5 The function is negative when it is below the x axis.
34 and
2x x
The quadratic formula2 0ax bx c
2 4
2
b b acx
a
21. Solve 2 5 1 0x x 2compare with 0ax bx c 2, 5, 1a b c
5 25 8
4x
5 17 5 17 and
4 4
2.28 and 0.22
22. Solve 6 9 0x x 2compare with 0ax bx c 1, 6, 9a b c
6 36 36
2x
6 0
2
3
From the above example when the number under the square root sign is zero there is only 1 solution.
23. Solve 2 9 0x x 2compare with 0ax bx c 1, 2, 9a b c
2 4 36
2x
2 32
2
Since 32 is not a real number there are no real roots.
From the above example we require the number under the square root sign to be positive in order for 2 real roots to exist.
This leads to the following observation.
2If 4 0 the roots are real and unequal.b ac 2If 4 0 the roots are real and equal.b ac 2If 4 0 the roots are non-real.b ac
2
2
For the quadratic equation 0,
4 is called the DISCRIMINANT.
ax bx c
b ac
24. Find the nature of the roots of 4 12 9 0x x
4, 12, 9a b c 2 4 144 144b ac
0
Since the discriminant is zero, the roots are real and equal.
Using the discriminantWe can use the discriminant to find unknown coefficients in a quadratic equation.
21. Find given that 2 4 0 has real roots.p x x p
2, 4,a b c p 2 4 16 8b ac p 0
8 16p2p
The equation has real roots when 2.p
23. Show that the roots of ( 2) (3 2) 2 0 are always real. k x k x k
( 2), (3 2), 2a k b k c k
2 24 (4 12 9 ) 8 ( 2)b ac k k k k
2 24 12 9 8 16k k k k 2 4 4k k
2( 2)k 2( 2) 0 for all values of .k k
2 3k
Since the discriminant is always greater than or equal to zero, the roots of the equation are always real.
Conditions for tangencyTo determine whether a straight line cuts, touches or does not meet a curve the equation of the line is substituted into the equation of the curve.
When a quadratic equation results, the discriminant can be used to find the number of points of intersection.
2If 4 0 there are two distinct points of intersection.b ac
2If 4 0 there is only one point of intersection.
the line is a tangent to the curve.
b ac
2If 4 0 the line does not intersect the curve.b ac
21. Prove that the line 2 1 is a tangent to the curve .
Find the point of intersection.
y x y x
2 2 1x x 2 2 1 0x x
1, 2, 1a b c 2 4 4 4b ac 0
Since the discriminant is zero, the line is a tangent to the curve.
2 2 1 0x x 2( 1) 0x
1x 2 1y x
Hence the point of intersection is (1 , 1).
22. Find the equation of the tangent to 1 that has gradient 2. y x
A straight line with gradient 2 is of the form 2y x k 2 1 2x x k
1, 2, 1a b c k
2For tangency 4 0b ac 4 4(1 ) 0k 4 4 4 0k
4 0k 0k
Hence the equation of the tangent is y = 2x.
2 2 (1 ) 0x x k
23. Find the equations of the tangents from (0,-2) to the curve 8 . y x
A straight line passing through (0,-2) is of the form 2y mx 28 2x mx
8, , 2a b m c 2For tangency 4 0b ac 2 64 0m
2 64m 8m
Hence the equation of the two tangents are y = 8x – 2 and y = -8x - 2.
28 2 0x mx y
x