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