4.4 Different Forms of Quadratic Expressions Secondary Math 2 4.4 Different Forms of Quadratic Expressions

Warm Up Use any method to solve. 𝑥 2 −13𝑥−68=0

𝑥 2 −13𝑥−68=0

𝑥 2 −13𝑥−68=0

𝑥 2 −13𝑥−68=0

Recap of this unit so far… We can solve quadratic equations using three different methods: Factoring Quadratic Formula Completing the Square

What you will learn The three different ways we can write a quadratic expression, and what they mean.

Standard Form: y=𝑎 𝑥 2 +𝑏𝑥+𝑐 Vertex Form: 𝑦=𝑎 𝑥−ℎ 2 +𝑘 Factored Form: 𝑦=𝑎(𝑥+𝑒)(𝑥+𝑓)

y-intercept: (0, c) Standard Form Standard form: 𝑓 𝑥 =𝑎 𝑥 2 +𝑏𝑥+𝑐 It is very easy to find the y-intercept from the standard form. y-intercept: (0, c)

Factored Form If we are interested in finding the x-intercepts of a quadratic, the factored form can help us find this. 𝑓 𝒙 =𝒂(𝒙+𝒆)(𝒙+𝒇) has x-intercepts at −𝑒, 0 , 𝑎𝑛𝑑 (−𝑓, 0).

𝑓 𝑥 =𝑎 𝑥−ℎ 2 +𝑘 The vertex of the parabola will be at 𝒉, 𝒌 . Vertex Form 𝑓 𝑥 =𝑎 𝑥−ℎ 2 +𝑘 The vertex of the parabola will be at 𝒉, 𝒌 .

Example y-intercept: (0, 12) Consider the quadratic 𝑓 𝑥 =− 𝑥 2 −4𝑥+12 Factored form: 𝑓 𝑥 =−(𝑥+6)(𝑥−2) Vertex form: 𝑓 𝑥 =− 𝑥+2 2 +16 vertex: (−2, 16) 𝑥-intercepts: (-6,0), (2,0)

Example Consider the quadratic 𝑓 𝑥 = 𝑥 2 −10𝑥+16 Factored form: 𝑓 𝑥 = 𝑥−8 𝑥−2 Vertex form: 𝑓 𝑥 = 𝑥−5 2 −9 𝑥-intercepts: (2,0), (8,0) vertex: (5, −9)

1) 36 𝑎 2 −1=63

2) 25 𝑟 2 −5=76

3) 𝑛 2 +40=−13𝑛

4) 𝑛 2 =3𝑛

5) 9 𝑥 2 −19=0

6) 11 𝑟 2 −2𝑟=6

7) 𝑝 2 −10𝑝−24=0

8) 𝑚 2 −14𝑚+33=0

Consider the following quadratic. 𝑥 2 +12𝑥+32=0 Exit Problem Consider the following quadratic. 𝑥 2 +12𝑥+32=0 Solve it two different ways. (Factor, complete the square, or quadratic formula).