1. Drawing Quadratic Curves – Part 2
Slideshow 28, Mathematics
Mr. Richard Sasaki

2. Objectives
Learn how to write an equation 𝑦=𝑎 𝑥 2 +𝑏𝑥+𝑐 in the form 𝑦=𝑎 𝑥−ℎ 2 +𝑘
Be able to draw graphs in the form 𝑦=𝑎 𝑥 2 +𝑏𝑥+𝑐 when considering the vertex (ℎ, 𝑘)

3. Algebra Manipulation
If we write 𝑦=𝑎 𝑥 2 +𝑏𝑥+𝑐 in the form 𝑦=𝑎 𝑥−ℎ 2 +𝑘 (vertex form), we know what the vertex is because its coordinates are in the form .
(ℎ, 𝑘)
Don’t remove the factor of 𝑥 too!
Example
Write 𝑦=2 𝑥 2 +16𝑥+30 in vertex form.
𝑦−30=2 𝑥 2 +16𝑥
1. Subtract both sides by 30.
𝑦−30=2( 𝑥 2 +8𝑥)
2. Factorise the right by the 𝑥 2 coefficient.
1 2 ∙8 2 = 16
3. Halve and then square the 𝑥 coefficient.
4. Add and subtract 16 to produce a perfect square.
𝑦−30=2( 𝑥 2 +8𝑥+16−16)
𝑦−30=2 𝑥+4 2 −16
𝑦−30=2 ∙ 𝑥+4 2 −2∙16
5. Expand the outer bracket and rewrite.
𝑦−30=2 𝑥+4 2 −32
⇒𝑦=2 𝑥+4 2 −2

4. Algebra Manipulation Let’s do one more example! Example
Write 𝑦=3 𝑥 2 −30𝑥+78 in vertex form and write down its vertex.
𝑦−78=3 𝑥 2 −30𝑥
1. Subtract both sides by 78.
𝑦−78=3( 𝑥 2 −10𝑥)
2. Factorise the right by the 𝑥 2 coefficient.
3. Calculate half of the 𝑥 coefficient, then square it.
1 2 ∙10 2 = 25
4. Add and subtract 25 to produce a perfect square.
𝑦−78=3( 𝑥 2 −10𝑥+25−25)
𝑦−78=3 𝑥−5 2 −25
5. Expand the outer bracket and rewrite.
𝑦−78=3 ∙ 𝑥−5 2 −3∙25
𝑦−78=3 𝑥−5 2 −75
⇒𝑦=3 𝑥−
The vertex is at .
(5, 3)

5. 𝑦=4 𝑥 2 −3
𝑦=6 𝑥 2 +12𝑥+3
𝑦=4 𝑥 2 +16𝑥+15
𝑦=−2 𝑥 2 −12𝑥−22
𝑦=3 𝑥
𝑦=2 𝑥+5 2 −1
𝑦= 𝑥+4 2 −3
𝑦= 4 𝑥−
𝑦= 6 𝑥−1 2 −9
𝑦= 2 𝑥+2 2 −7

6. Answers - Hard 𝑦=4 𝑥−6 2 +3 𝑦=− 𝑥+1 2 +2 𝑦=2 𝑥+8 2 −52 (6, 3) (−1, 2)
𝑦=4 𝑥−
𝑦=− 𝑥
𝑦=2 𝑥+8 2 −52
(6, 3)
(−1, 2)
(−8, −52)
Minimum
Maximum
Minimum
𝑦=3 𝑥−11 2 −100
𝑦=−2 𝑥
𝑦=7 𝑥−5 2 −210
(11, −100)
(−5, 20)
(5, −210)
Minimum
Maximum
Minimum
𝑦=−3 𝑥 −2
𝑦=2 𝑥−
(0.5, 3)
(− 1 2 , −2)
Minimum
Maximum

7. Drawing Parabolae from Quadratics
If we are able to write a quadratic 𝑦=𝑎 𝑥 2 +𝑏𝑥+𝑐 in the form 𝑦=𝑎 𝑥−ℎ 2 +𝑘, we can easily draw it!
Example
Write 𝑦=2 𝑥 2 −4𝑥+4 in vertex form and hence, draw it.
𝑦−4=2 𝑥 2 −4𝑥 −1 1 3 𝒙 𝑦 2
𝑦−4=2( 𝑥 2 −2𝑥) 10 4 2 4 10
𝑦−4=2( 𝑥 2 −2𝑥+1−1)
𝑦−4=2( 𝑥−1 2 −1)
We call this the parabola’s .
𝑦−4=2 𝑥−1 2 −2
𝑦=2 𝑥−
The line takes place at
axis of symmetry
𝑥=ℎ
Vertex:
(1, 2)

8. Answers (Easy)
𝑦=2 𝑥− 2 3
𝑦= 𝑥−6 2 −5 6 −5

9. Answers (Hard)
𝑦= 𝑥− −2 2 −2
𝑦=− 𝑥− −8 4 −8

10. Drawing Parabolae from Quadratics
Do you know another way of finding the parabola 𝑦=𝑎 𝑥 2 +𝑏𝑥+𝑐?
Let’s write the general form in vertex form.
𝑦=𝑎 𝑥+ 𝑏 2𝑎 2 − 𝑏 2 4𝑎 +𝑐
𝑦−𝑐=𝑎 𝑥 2 +𝑏𝑥
𝑦−𝑐=𝑎 𝑥 2 + 𝑏𝑥 𝑎
𝑦=𝑎 𝑥+ 𝑏 2𝑎 2 − 𝑏 2 4𝑎 + 4𝑎𝑐 4𝑎
𝑦−𝑐=𝑎 𝑥 2 + 𝑏𝑥 𝑎 + 𝑏 2𝑎 2 − 𝑏 2𝑎 2
𝑦=𝑎 𝑥− − 𝑏 2𝑎 𝑎𝑐− 𝑏 2 4𝑎
𝑦−𝑐=𝑎 𝑥+ 𝑏 2𝑎 2 − 𝑏 2 4 𝑎 2
− 𝑏 2𝑎
4𝑎𝑐− 𝑏 2 4𝑎
Vertex: ( , )
𝑦−𝑐=𝑎 𝑥+ 𝑏 2𝑎 2 −𝑎 𝑏 2 4 𝑎 2
ℎ=− 𝑏 2𝑎 , 𝑘= 4𝑎𝑐− 𝑏 2 4𝑎

11. Using the Formulae
Now we know that ℎ=− 𝑏 2𝑎 and 𝑘= 4𝑎𝑐− 𝑏 2 4𝑎 , we can use these to find the vertex for an expression 𝑦=𝑎 𝑥 2 +𝑏𝑥+𝑐 without having to write it in vertex form.
Example
Calculate where the vertex takes place for 𝑦=2 𝑥 2 −4𝑥+5 without writing the expression in vertex form.
ℎ=− 𝑏 2𝑎
=− −4 2∙2 =1
𝑘= 4𝑎𝑐− 𝑏 2 4𝑎
= 4∙2∙5− (−4) 2 4∙2
= 40−16 8 =3
∴ Vertex: (1, 3)

12. (2, −2)
(5, 6)
(3, 2)
(−1,−8)
(8, 7)
(−3,− 6)
(1, 18)
(4, −8)
(9, 6)
(25, −10)

13. Answers
(4, −1)
(−6, 1)