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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)
@welshfella @Wxlfz @foreverthebeast @triciaal @ganeshie8 @campbell_st
@dan815 @jim_thompson5910 @jagr2713 @Nnesha @undeadknight26
vertex form a(x-h)^2 + k h = -b/(2a) k = f(h)
vertex (h, k)
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
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
Where did you get 12t from?
It;s not making sense.
what do you mean? look at the original
In both the second t disappears.
Ok, so here we go.
you have a parabola t^2 + 12t -18 to find the zeros set = 0 and solve
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
What did I do wrong?
No writing, just type please.
4 th line is wrong and set = 0
ok, what needs to happen in line 4?
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
@jim_thompson5910 can you continue please I have to leave
@Nnesha @dan815 I have to leave
@dan815 @jim_thompson5910 @undeadknight26 @robtobey @Elsa213 @zzr0ck3r
now take (t + 6)^2 - 36 − 18 and simplify to get (t + 6)^2 - 54
compare (t + 6)^2 - 54 to the form a(x-h)^2 + k in this case, 't' takes the place of 'x'
the vertex of y = a(x-h)^2 + k is (h,k)
Ok, well, t needs to go outside of the parentheses.
what do you mean?
Well, it's no the same form, the only difference is the t (or a) being outside of the parentheses.
think of (t + 6)^2 - 54 as 1(t + 6)^2 - 54
MISSING t (t + 6)^2 - 54 to the form a(x-h)^2 + k
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
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
yeah it looks good
what's the vertex of (t + 6)^2 - 54
Not, sure, Mathway couldn't calculate an answer.
Do you see how I wrote (x+6)^2 - 54 into 1(x - (-6))^2 - 54 ?
yes that is the vertex
Ok, was is is and why?
Determine the vertex and indicate whether it is a maximum or a minimum on the graph. How do you know?
Vertex: (-6, -54), this is a minimum because because the parabola is pointing down.
is that right?
yeah it looks like this |dw:1438291343433:dw|
Part C: Axis of symmetry is -6.
more like x = -6, but yeah