anonymous
  • anonymous
Help Would Be Greatly Appreciated.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The function f(t) = t2 + 12t − 18 represents a parabola. Part A: Rewrite the function in vertex form by completing the square. Show your work. (6 points) Part B: Determine the vertex and indicate whether it is a maximum or a minimum on the graph. How do you know? (2 points) Part C: Determine the axis of symmetry for f(t). (2 points)
anonymous
  • anonymous
@welshfella @Wxlfz @foreverthebeast @triciaal @ganeshie8 @campbell_st
anonymous
  • anonymous
@dan815 @jim_thompson5910 @jagr2713 @Nnesha @undeadknight26

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triciaal
  • triciaal
vertex form a(x-h)^2 + k h = -b/(2a) k = f(h)
anonymous
  • anonymous
I'm lost.
triciaal
  • triciaal
|dw:1438288504884:dw|
triciaal
  • triciaal
vertex (h, k)
triciaal
  • triciaal
the vertex is the point with maximum or minimum y value when a is positive then y is positive and the vertex is a maximum when a is negative then y is negative and the vertex is at the minimum
triciaal
  • triciaal
|dw:1438289057781:dw|
anonymous
  • anonymous
Part A: f(t) = t^2 + 12t − 18 f(t) = (t + 12/2)^2 - (12/2)^2 − 18 f(t) = (t + 6)^2 - 36 − 18 f(t) = (t + 36) - 36 − 18 f(t) = t − 18
anonymous
  • anonymous
@triciaal
triciaal
  • triciaal
|dw:1438289240219:dw|
anonymous
  • anonymous
Where did you get 12t from?
anonymous
  • anonymous
It;s not making sense.
triciaal
  • triciaal
what do you mean? look at the original
anonymous
  • anonymous
In both the second t disappears.
anonymous
  • anonymous
Ok, so here we go.
triciaal
  • triciaal
you have a parabola t^2 + 12t -18 to find the zeros set = 0 and solve
anonymous
  • anonymous
f(t) = t^2 + 12t − 18 f(t) = (t + 12/2)^2 - (12/2)^2 − 18 f(t) = (t + 6)^2 - 36 − 18 f(t) = (t^2 + 36) - 36 − 18 f(t) = t^2 − 18
anonymous
  • anonymous
What did I do wrong?
anonymous
  • anonymous
No writing, just type please.
triciaal
  • triciaal
4 th line is wrong and set = 0
anonymous
  • anonymous
ok, what needs to happen in line 4?
triciaal
  • triciaal
|dw:1438289870837:dw|
anonymous
  • anonymous
Part A: t^2 + 12t − 18 = 0 (t + 12/2)^2 - (12/2)^2 − 18 = 0 (t + 6)^2 - 36 − 18 = 0 (t + 6)^2 = t^2 + 12t + 36
triciaal
  • triciaal
@jim_thompson5910 can you continue please I have to leave
triciaal
  • triciaal
@Nnesha @dan815 I have to leave
anonymous
  • anonymous
@dan815 @jim_thompson5910 @undeadknight26 @robtobey @Elsa213 @zzr0ck3r
jim_thompson5910
  • jim_thompson5910
now take (t + 6)^2 - 36 − 18 and simplify to get (t + 6)^2 - 54
jim_thompson5910
  • jim_thompson5910
compare (t + 6)^2 - 54 to the form a(x-h)^2 + k in this case, 't' takes the place of 'x'
jim_thompson5910
  • jim_thompson5910
the vertex of y = a(x-h)^2 + k is (h,k)
anonymous
  • anonymous
Ok, well, t needs to go outside of the parentheses.
jim_thompson5910
  • jim_thompson5910
what do you mean?
anonymous
  • anonymous
Well, it's no the same form, the only difference is the t (or a) being outside of the parentheses.
jim_thompson5910
  • jim_thompson5910
think of (t + 6)^2 - 54 as 1(t + 6)^2 - 54
anonymous
  • anonymous
MISSING t (t + 6)^2 - 54 to the form a(x-h)^2 + k
anonymous
  • anonymous
Ahh, ok.
jim_thompson5910
  • jim_thompson5910
I'm going to replace t with x 1(x + 6)^2 - 54 1(x - (-6))^2 - 54 The expression 1(x - (-6))^2 - 54 is in the form a(x-h)^2 + k a = 1 h = -6 k = -54
anonymous
  • anonymous
Ok
anonymous
  • anonymous
Is this correct? Part A: t^2 + 12t − 18 = 0 (t + 12/2)^2 - (12/2)^2 − 18 = 0 (t + 6)^2 - 36 − 18 = 0 (t + 6)^2 - 54
jim_thompson5910
  • jim_thompson5910
yeah it looks good
jim_thompson5910
  • jim_thompson5910
what's the vertex of (t + 6)^2 - 54
anonymous
  • anonymous
Not, sure, Mathway couldn't calculate an answer.
anonymous
  • anonymous
-54?
jim_thompson5910
  • jim_thompson5910
Do you see how I wrote (x+6)^2 - 54 into 1(x - (-6))^2 - 54 ?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
(-6, -54)
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
yes that is the vertex
anonymous
  • anonymous
Ok, was is is and why?
jim_thompson5910
  • jim_thompson5910
huh?
anonymous
  • anonymous
Determine the vertex and indicate whether it is a maximum or a minimum on the graph. How do you know?
anonymous
  • anonymous
Vertex: (-6, -54), this is a minimum because because the parabola is pointing down.
anonymous
  • anonymous
is that right?
jim_thompson5910
  • jim_thompson5910
yeah it looks like this |dw:1438291343433:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1438291350542:dw|
anonymous
  • anonymous
Part C: Axis of symmetry is -6.
jim_thompson5910
  • jim_thompson5910
more like x = -6, but yeah
triciaal
  • triciaal
@jim_thompson5910 thanks

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