Sketching quadratic functions
To sketch a quadratic function we need to identify where possible:
The shape:
ax 2  bx  c
If a  0 then
If a  0 then
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.
1. Sketch the graph of y  5  4 x  x 2
The shape
10
2
4
6
8
– 10
2
4
6
8
2– 10
4
6
8
10
2
4
6
8
The coefficient of x2 is -1 so the shape is
The Y intercept
(0 , 5)
y
(-2 , 9)
10
8
The roots
6
5  4x  x  0
(5  x)(1  x)  0
2
4
2
– 10 – 8
(-5 , 0) (1 , 0)
– 6
– 4
– 2
– 2
– 4
The axis of symmetry
– 6
– 8
Mid way between -5 and 1 is -2
x = -2
The coordinates of the turning point
When x  2, y  9
(-2 , 9)
– 10
2
4
6
8
10
x
Completing the square
The 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
y  ax 2  bx  c can be written in the form y  a( x  p)2 +q.
Axis of symmetry x   p
Coordinates of turning point ( p, q).
1. Find the equation of the axis of symmetry and the coordinates of the
turning point of y  2 x 2  8 x  9.
y  2 x2  8x  9
 2( x 2  4 x)  9
 2( x  2)2  9  8
 2( x  2)2  1
Axis of symmetry is x = 2
Coordinates of the minimum turning point is (2 , 1)
2. Find the equation of the axis of symmetry and the coordinates of the
turning point of y  7  6 x  x 2 .
y   x2  6x  7
 ( x 2  6 x)  7
 ( x  3)2  7  9
 ( x  3)2  16
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
1. Solve 6 x 2  x  15  0
(2 x  3)(3x  5)  0
2x  3  0
3
x
2
3x  5  0
5
x
3
2. Solve x 2  4 x  1  0
This does not factorise.
ax 2  bx  c  0
a  1, b  4, c  1
4  16  4
x
2
4  20

2
42 5

2
2(2  5)

2
 2  5 or 2  5
b  b 2  4ac
x
2a
Quadratic inequations
A quadratic inequation can be solved by using a sketch of the quadratic function.
1. For what values of x is 12  5 x  2 x 2  0?
First do a quick sketch of the graph of the function.
12  5 x  2 x 2  (4  x)(3  2 x)
Roots are -4 and 1.5
Shape is a
-4
1.5
The function is positive when it is above the x axis.
3
4  x 
2
2. For what values of x is 12  5 x  2 x 2  0?
First do a quick sketch of the graph of the function.
12  5 x  2 x 2  (4  x)(3  2 x)
Roots are -4 and 1.5
Shape is a
-4
1.5
The function is negative when it is below the x axis.
3
4  x and x 
2
The quadratic formula
ax 2  bx  c  0
b  b 2  4ac
x
2a
2
1. Solve 2 x 2  5x  1  0 compare with ax  bx  c  0
a  2, b  5, c  1
5  25  8
x
4
5  17
5  17

and
4
4
 2.28 and 0.22
2. Solve x 2  6 x  9  0
a  1, b  6, c  9
compare with ax 2  bx  c  0
6  36  36
x
2
60

2
3
From the above example when the number under the square root sign is zero there
is only 1 solution.
3. Solve x 2  2 x  9  0
a  1, b  2, c  9
compare with ax 2  bx  c  0
2  4  36
x
2
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.
If b2  4ac  0 the roots are real and unequal.
If b2  4ac  0 the roots are real and equal.
If b2  4ac  0 the roots are non-real.
For the quadratic equation ax 2  bx  c  0,
b 2  4ac is called the DISCRIMINANT.
4. Find the nature of the roots of 4 x 2  12 x  9  0
a  4, b  12, c  9
b2  4ac  144  144
0
Since the discriminant is zero, the roots are real and equal.
Using the discriminant
We can use the discriminant to find unknown coefficients in a quadratic equation.
1. Find p given that 2 x 2  4 x  p  0 has real roots.
a  2, b  4, c  p
b2  4ac  16  8 p  0
8 p  16
p2
The equation has real roots when p  2.
3. Show that the roots of (k  2) x 2  (3k  2) x  2k  0 are always real.
a  (k  2), b  (3k  2), c  2k
 2  3k
b2  4ac  (4  12k  9k 2 )  8k (k  2)
 4  12k  9k 2  8k 2  16k
 k 2  4k  4
 (k  2)2
(k  2)2  0 for all values of k.
Since the discriminant is always greater than or equal to zero, the roots of the
equation are always real.
Conditions for tangency
To 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.
If b2  4ac  0 there are two distinct points of intersection.
If b 2  4ac  0 there is only one point of intersection.
the line is a tangent to the curve.
If b2  4ac  0 the line does not intersect the curve.
1. Prove that the line y  2 x  1 is a tangent to the curve y  x 2 .
Find the point of intersection.
x2  2 x 1
x2  2 x  1  0
a  1, b  2, c  1
b2  4ac  4  4  0
Since the discriminant is zero, the line is a tangent to the curve.
x2  2 x  1  0
( x  1)2  0
x 1
y  x2  1
Hence the point of intersection is (1 , 1).
2. Find the equation of the tangent to y  x 2  1 that has gradient 2.
A straight line with gradient 2 is of the form y  2 x  k
x2  1  2 x  k
x 2  2 x  (1  k )  0
a  1, b  2, c  1  k
For tangency b2  4ac  0
4  4(1  k )  0
4  4  4k  0
4k  0
k 0
Hence the equation of the tangent is y = 2x.
3. Find the equations of the tangents from (0,-2) to the curve y  8x 2 .
A straight line passing through (0,-2) is of the form y  mx  2
8 x 2  mx  2
y
8 x 2  mx  2  0
a  8, b  m, c  2
For tangency b2  4ac  0
x
m2  64  0
m2  64
m  8
Hence the equation of the two tangents are y = 8x – 2 and y = -8x - 2.