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anastassius [24]
3 years ago
14

The function f(t) = 4t^2 − 8t + 6 shows the height from the ground f(t), in meters, of a roller coaster car at different times t

. Write f(t) in the vertex form a(x − h)^2 + k, where a, h, and k are integers, and interpret the vertex of f(t).
A - f(t) = 4(t − 1)^2 + 3; the minimum height of the roller coaster is 3 meters from the ground
B - f(t) = 4(t − 1)^2 + 3; the minimum height of the roller coaster is 1 meter from the ground
C - f(t) = 4(t − 1)^2 + 2; the minimum height of the roller coaster is 2 meters from the ground
D - f(t) = 4(t − 1)^2 + 2; the minimum height of the roller coaster is 1 meter from the ground
Mathematics
1 answer:
andreev551 [17]3 years ago
3 0

Answer:

C - f(t) = 4(t − 1)^2 + 2; the minimum height of the roller coaster is 2 meters from the ground.

Step-by-step explanation:

Here we're asked to rewrite the given equation f(t) = 4t^2 − 8t + 6 in the form f(t) = a(t - h)^2 + k (which is known as the "vertex form of the equation of a parabola.")  Here (h, k) is the vertex and a is a scale factor.

Let's begin by factoring 4 out of all three terms:

f(t) = 4 [ t^2 - 2t + 6/4 ]

Next, we must "complete the square" of t^2 - 2t + 6/4; in other words, we must re-write this expression in the form (t - h)^2 + k.

(To be continued)

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It is given that two vertices of square are (0,0) and (4,2).

Now the problem is that you haven't given that whether these two vertices are adjacent vertices or opposite vertices of the square.

1. By Supposing that these two are adjacent vertices of Square

The third vertex will be at (-4,2) which lies in third quadrant.

Suppose the coordinate of fourth vertex be (x,y).

Mid point of line joining (4,2) and (-4,2) is{ [4+(-4)]/2,(2+2)/2} is (0,2).

Mid point of line joining (x,y) and (0,0) is (x/2,y/2).

Since diagonals of square bisect each other,

∵ x/2=0

⇒x=0

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y/2=2

⇒y=4

So, The Coordinate of  fourth vertex is (0,4).

Now coming back to second condition if these are two opposite vertex of Square.

Let the third coordinate be (a,b).

Length of diagonal=\sqrt{(4-0)^2+(2-0)^2}=\sqrt20=2\sqrt5

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Then length of Diagonal of square =√2 A

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As third vertex is (a,b).

Using distance formula

a² + b²=10  -------------(1)

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Solving expression (1) and (2), we get

⇒a²+ b²=(a-4)² +(b-2)²

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Putting the value of b in (1),we get

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Splitting the middle term,we get

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⇒a=3  ∧  a=1

we get b=5-2×1=3 and b=5-2×3=5-6=-1

So,the other vertex are (1,3) and(3,-1).





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