TURNING POINTS OF POLYNOMIALS
Let us imagine traveling along the graph of a polynomial, moving from left to right. Sometimes we go ‘uphill’,
sometimes ‘downhill’, and sometimes we change direction. Such a change of direction is called a turning point.
The purpose of this section is to make this concept more precise
Roughly, a maximum is a point on a curve where there is a ‘ high spot’.
That is, if we ‘stand on’ the point and don't look too far away,
then everything we see is lower (or at the same height).
If we look too far away ( at right) then we may see points that are higher.
People often say ‘ max’ instead of ‘ maximum’, for brevity.
There is same description for a minimum (‘ min’, for short).
DEFINITION - Turning points of polynomials:
A turning point of a polynomial is a point where there is a max or a min.
Notes about Turning Points:
At a max, we stop going up, and start going down.
At a min, we stop going down, and start going up.
For polynomials, a max or min always occurs at a horizontal tangent line.
Thus, a turning point of a polynomial always occurs at a horizontal tangent line.
It's possible to have a horizontal tangent line on a polynomial that is not a turning point, as shown below.
At the black point below, we don't change direction:
we travel uphill to the left of the black point;
we also travel uphill to the right of the black point.
Prob :Given that 𝑥 2 − 5𝑥 + 13 ≡ 𝑥 − 𝑝 2 + 𝑞 for all values of 𝑥 , calculate the value of 𝑝 and of 𝑞 .Hence state
a) the minimum value of 𝑥 2 − 5𝑥 + 13,
b) the value of 𝑥 at which the minimum value occurs,
c)Sketch the curve of 𝑥 2 − 5𝑥 + 13.
5
5
5
Ans. 𝑥 2 − 5𝑥 + 13 = 𝑥 2 − 2 × 𝑥×2 + (2)2 − (2)2 + 13
5
= (𝑥 − 2)2 −
5
3
25
4
+ 13
= (𝑥 − 2)2 +6 4 ≡ 𝑥 − 𝑝
5
2
+𝑞
3
𝑝 = 2 and 𝑞 = 6 4
5
a) the minimum value of 𝑥 2 − 5𝑥 + 13 occurs when (𝑥 − 2)2 = 0
5
3
If (𝑥 − 2)2 = 0 , the minimum value of 𝑥 2 − 5𝑥 + 13 is 6 4 .
5
b) (𝑥 − 2)2 =0
5
𝑥 − 2 =0
5
so, 𝑥 = 2
c) Graphical presentation:
HW
Marks-10
Prob. Find the maximum value of −2𝑥 2 + 4𝑥 − 5 and the value of 𝑥 at which the maximum value
occurs .
Sketch the curve 𝑦 = −2𝑥 2 + 4𝑥 − 5
−2𝑥 2 + 4𝑥 − 5 = −2(𝑥 2 − 2𝑥) − 5
= −2(𝑥 2 − 2 × 𝑥 × 1 − 12 )−5
= −2 𝑥 − 1 2 + 2 − 5
= −2 𝑥 − 1 2 − 3
Maximum value is -3 and it occurs when 𝑥 − 1 2 =0
𝑥−1=0
i.e. 𝑥 = 1

Graphical presentation:
Assessment
Total Marks-10
Time-15 minutes
Prob 1.Express 𝑦 = 4𝑥 2 − 6𝑥 + 14 in the form 𝑦 = 4(𝑥 − 𝑝)2 +𝑞 where 𝑝 and 𝑞 are constants
and hence find the minimum value of 𝑦 and the value of x at which this occurs .
Sketch the curve 𝑦 = 4𝑥 2 − 6𝑥 + 14
Answer of the assignment
𝑦 = 4𝑥 2 − 6𝑥 + 14
6
= 4(𝑥 2 − 𝑥) +14
4
3
= 4(𝑥 2 − 𝑥) +14
2
3 2
3
= 4 (𝑥 − ) −( )2 +14
4
4
3
4
= 4(𝑥 − )2 −4 ×
9
16
3
9
4
4
3 2
3
) +11
4
4
+ 14
= 4(𝑥 − )2 − + 14
= 4(𝑥 −
This lesson is prepared by Manik Lal Roy