## anonymous one year ago Find the vertex of the graph of the function. f(x) = 4x^2 + 24x + 32

1. Jhannybean

Basically you're taking your equation $$y=4x^2+24+32$$ that has the quadratic form $$y=ax^2+bx+c$$ and converting it into vertex form: $$y=a(x-h)^2+k$$

2. Jhannybean

So to start, we can immediately see that all the coefficients are divisible by 4. So simplify your function.

3. anonymous

y=4x2+24+32= y=ax2+bx+c ? = y=a(x−h)2+k right?

4. anonymous

a=4 b=24 c=32

5. anonymous

y=a(x−h)2+k = = y=4(x−h)2+k

6. Jhannybean

You got the idea somewhat, but first lets simplify this :) You'll see what I mean by "fitting the equation into the form" :D

7. anonymous

yeah, kinda. what i don't get is how to get x&y values to then find h&k

8. anonymous

ohh and thank you so much for helping me @Jhannybean

9. Jhannybean

Simplify this: $\large y=\frac{4}{4}x^2+\frac{24}{4}x+\frac{32}{4}$

10. Jhannybean

@DanTheMan99

11. anonymous

thank you

12. anonymous

sorry mi wifi was acting up @Jhannybean

13. anonymous

y=x2+6x+8

14. anonymous

y=x2+6x+8 y=ax2+bx+c ? y=a(x−h)2+k

15. anonymous

but how will i get the other values?

16. Jhannybean

Alright, now that we simplified it to $$y=x^2+6x+8$$, we need to use the method of completing the square. Have you learned that method yet?

17. anonymous

not yet

18. Jhannybean

In the method of completing the square, we will basically create another quadratic function into the quadratic function we already have, $$y=x^2+6x+8$$ It's like an inception of quadratics that will help you find the vertex. :P

19. Jhannybean

So let's work it out slowly. First set your function equal to zero. $$y=x^2+6x+8=0$$ Next, move over your constants to one side of the equation, while leaving your variables on the other side side. $$y=x^2+6x=-8$$ Following so far?

20. anonymous

yes

21. anonymous

do i get x and y intercepts?

22. Jhannybean

Once we're dne you'll have your x and y intercepts and you'll see what I mean, just follow me on this one :P Trust!

23. Jhannybean

Now we're going to create a quadratic on the left hand side of the function by completing the square of the left hand side. Essentially we're going to finish the quadratic on the left hand side so it resembles the form $$ax^2+bx+c$$ but right now, we've only got $$ax^2+bx$$ and THEREFORE need to find ourselves a new $$c$$ value.

24. Jhannybean

$c=\left(\frac{b}{2}\right)^2 \longrightarrow c=\left(\frac{6}{2}\right)^2 = (3)^2 = 9$

25. anonymous

so c = 9

26. Jhannybean

Yep. So since we're adding c to the left side, we have to respectively add it to the right side as well

27. Jhannybean

$y=x^2+6x\color{red}{+9}=-8\color{red}{+9}$

28. anonymous

y=x2+6x+9=−8+9 y=x2+6x+9=+1

29. anonymous

ohh ok

30. Jhannybean

If you're good with simplifying quadratics, you'll see that the left hand side now is a PERFECT square! :o $$x^2+6x+9 \longrightarrow (x+3)^2$$

31. Jhannybean

Rewrite it. $y=(x+3)^2=1$

32. Jhannybean

Well, you would move the 1 over to fit your new equation to fit the vertex form $$y=a(x+h)^2+k$$

33. Jhannybean

Therefore $$y=a(x+h)^2+k \longrightarrow y=(x+3)^2-1$$

34. Jhannybean

Do you see how it works?

35. Jhannybean

Whoop, I wrote it wrong, $$y=a(x-h)^2+k$$** Typo.

36. anonymous

y=a(x−h)2+k⟶y=(x+3)2−1

37. anonymous

⟶y=(x+3)^2−1

38. Jhannybean

So we know a positive can be split into 2 negatives, therefore $$y=a(x-h)^2+k \longrightarrow y=(x-(-3))^2-1$$

39. Jhannybean

So tell me, what are your x and y intercepts? :D

40. anonymous

y=-1

41. anonymous

xintercept = (x−(−3))2−1 = 1 = (x−(−3))2

42. Jhannybean

Im not understanding that.

43. anonymous

1 = (x−(−3))^2

44. Jhannybean

$y=a(x\color{red}{-h})^2\color{blue}{+k} \longrightarrow y=(x-\color{red}{(-3)})^2\color{blue}{-1}$

45. Jhannybean

46. anonymous

thats much better, sorry i was confused

47. anonymous

$y=a(x−h)^{2}+k⟶y=(x−(−3))^{2}−1$

48. anonymous

wait so what do i do now? the intercepts?

49. Jhannybean

You're toldto find the vertex of the graph, which I helped you find in the post above :)

50. anonymous

thank you @Jhannybean , sorry im a little heard-headed but i just reread it and i completely understand

51. Jhannybean

Awesome! Glad I could help :)

52. anonymous

53. anonymous

@Jhannybean sorry to bother you again

54. anonymous

does it mean that the answer would be "b"?

55. anonymous

im so confused :(

56. anonymous

use is formula....$-\frac{ b }{ 2a }$

57. anonymous

do i plug in a=1 and b=6??

58. anonymous

$y=ax^2+bx+c$

59. anonymous

y=(x−(−3))^2−1 -$\frac{ -3 }{ 1 }$

60. anonymous

-(-3/2)

61. anonymous

right?

62. anonymous

|dw:1438666779185:dw|

63. anonymous

3

64. anonymous

a and b is coefficient

65. anonymous

all the numberse

66. anonymous

24/8

67. anonymous

68. anonymous

-3

69. anonymous

yea

70. anonymous

71. anonymous

@saseal

72. anonymous

would it be the -3,-4 or -4,-3

73. anonymous

lowest point is obviously the y-axis

74. anonymous

im so lost

75. dan815

dantheman

76. dan815

lisnten to me, do u know how transformations work with functions

77. anonymous

GET A LIFE DAN SRS

78. anonymous

no

79. anonymous

sorry

80. dan815

do you know how transformations work??

81. anonymous

kind of

82. dan815

okay how about completing the square

83. anonymous

f(x-3) = f(x) shifted 3 to the right as example

84. anonymous

ok i get that

85. anonymous

if i am subtracing 3 away from what i am inputting at the time, then it must appear later by 3 units

86. dan815

f(x) = 4x^2 + 24x + 32 the point im trying to get across is we want to write this equation in the form f(x) = a(x-b)^2+c this means the vertex is at (b,c)

87. anonymous

i know its - 3

88. dan815

okay forget that BS for now

89. dan815

do you get why this equation has vertex at b,c though

90. anonymous

what i dont know is if it goes before or after the -4

91. dan815

f(x) = a(x-b)^2+c

92. dan815

no forget all that stuff, we are going to do this logically

93. dan815

f(x) = a(x-b)^2+c do you get why the vertex for this equation is b,c

94. dan815

95. dan815

y=x^2

96. dan815

how do you apply transformations to it?

97. dan815

y=(x-b)^2 would shift it b units to the right

98. dan815

y=(x-b)^2 + c would shift it b units to the right and c units up

99. dan815

you can think about it like this too (y-c)=(x-b)^2 y shifted c up and x shifted b right

100. anonymous

THIS MAN HAS CLEARLY LOST IT

101. anonymous

sorry my wifi wasn't working

102. anonymous

okay im following

103. dan815

do u get why the vertex for y=a(x-b)^2+c is at (b,c) then?

104. anonymous

yes

105. dan815

okay then y=4x^2 + 24x + 32 rewrite this in that form

106. anonymous

y=a(x-b)^2+c y=4(x-24)^2+32

107. dan815

thats not how it works

108. dan815

expand the brackets see if u get the right equation

109. anonymous

y=x^2+6x+8

110. anonymous

ohhhh

111. dan815

y=4(x-24)^2+32 =4*(x-24)(x-24)+32 = 4*(x^2-48x+24^2)+32

112. dan815

its much bigger than before now

113. dan815

so we have to transform to the form we want properly

114. anonymous

okay so you did the square

115. dan815

okay ill show u how to do it this time, u can try to do it later

116. anonymous

(a+b)^2=a^2+2ab+b^2

117. anonymous

= 4*(x^2-48x+24^2)+32

118. dan815

y=4x^2 + 24x + 32 y=4(x^2+6x)+32 ill complete the square in the brackets =4(x^2+6x+9-9)+32 =4((x^2+6x+9)-9)+32 =4*((x+3)^2-9)+32 =4*(x+3)^2+32-9*4 =4*(x+3)^2+32-36 =4*(x+3)^2-4 so the vertex is at (-3,-4)

119. anonymous

thank you so much

120. anonymous

uhhhhh finally i understand

121. dan815

y=4x^2 + 32x + 32 tell me the vertex for this equation and i will believe you

122. dan815

follow the steps i went through up there, its a very similar question

123. anonymous

@dan815 whoa you went thru all at

124. dan815

it looks like a lot of steps because im showing every little change, but normally once u learn the method u do it all in 1 or 2 steps

125. dan815

u can do these in ur head pretty much

126. anonymous

yea ikr thats why i said that lol

127. anonymous

y=4x^2 + 32x + 32 y=4(x^2+8x)+32 would it be that

128. anonymous

so i can continue

129. dan815

thats a start now u have to complete the square in those brackets go on

130. anonymous

y=4x^2 + 32x + 32 y=4(x^2+8x)+32 =4(x^2+8x+9-9)+32 =4((x^2+8x+9)-9)+32 =4*((x+4)^2-9)+32 =4*(x+4)^2+32-9*4 =4*(x+4)^2+32-36 =4*(x+4)^2-4

131. anonymous

:w

132. dan815

no

133. dan815

the question u are asking urself here is how can i get a perfect square out of x^2+8x

134. dan815

when u see x^2+8x you have to realize that the square has to be (x+4)^2 since this is the only square that will give u a x^2 and 8x now the constant is not there so we add and subtract that the constant part will be 4^2=16 so x^2+8x+16-16

135. anonymous

so you have to multiply 8 by 2

136. dan815

u are not multiplying 8 by 2

137. dan815

u are dividing (8/2) then squaring (8/2)^2 = 16 just so happesn that its also the same as 8*2

138. anonymous

y=4x^2 + 32x + 32 y=4(x^2+8x)+32 =4(x^2+8x+16-16)+32 =4((x^2+8x+16)-16)+32 =4*((x+4)^2-16)+32 =4*(x+4)^2+32-16*4

139. dan815

when u have x^2+ax and u want to write it in square form you take half of a as (x+a/2)^2=x^2+ax+(a/2)^2 ^--- term on x works out

140. dan815

yes now u got it so what is the vertex

141. anonymous

-4,-4??

142. dan815

no

143. dan815

what is the constant term

144. anonymous

16

145. dan815

=4*(x+4)^2+32-16*4 simplfy this to a(x-b)^2+c

146. dan815

=4*(x+4)^2+32-16*4 =4*(x+4)^2+32-64 =??

147. anonymous

-32

148. anonymous

=4*(x+4)^2-32

149. dan815

oaky so what is the vertex?

150. anonymous

i keep thinking its -4,-3

151. anonymous

but i don't think it is that

152. dan815

what is the vertex for this equation =a(x-b)^2+c

153. anonymous

b,c

154. dan815

okay so what is the vertex for this =4*(x+4)^2-32

155. anonymous

-4,-32

156. dan815

yes

157. anonymous

wow

158. dan815

:)

159. anonymous

that was really hard but now i got it lol

160. anonymous

only the inside changes sign, its the opposite

161. anonymous

thank you dan815

162. anonymous

you're a lifesaver!

163. dan815

no problem, try some completing the square excercises just to make sure though

164. dan815

or just do some expanding excercised, that should build your intuition up

165. anonymous

i tried doing it with analytic geometry method lol

166. anonymous

thank you

167. anonymous

works too

168. dan815

dantheman

169. dan815

for the record here is a short cut

170. anonymous

would it be the same for Writing the quadratic function in vertex form. y = x2 + 8x + 18

171. dan815

f(x) = 4x^2 + 24x + 32 you take the middle number and divide by -2* the first number 24/(-2*4) = -24/8=-3 now plug -3 into equation f(-3)=4*9+24*-3+32=-4 vertex is at (3,f(-3)) = (-3,-4)

172. anonymous

ohh wow that as much faster

173. dan815

NO do not use it

174. dan815

since u do not understand it

175. dan815

not until you understand completing the square method, then u can derive this

176. dan815

and then u can derive it one more time after u learn about derivatives

177. anonymous

yea i told him that stuff earlier lol -b/2a

178. dan815

f(x)=ax^2+bx+c prove that the vertex is at (-b/2a, ) x_coord=-b/2a y_coord=f(-b/2a) This can be your home work or something

179. anonymous

thank you

180. dan815

hint complete square for f(x)=ax^2+bx+c

181. anonymous

:)

182. anonymous

how can i rate a qualified helper

183. anonymous

(5 starts)

184. anonymous

lol

185. Jhannybean

Similar to my method without simplifying the equation first