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LUCKY_DIMON [66]
3 years ago
15

Use the drawing tool(s) to form the correct answers on the provided graph.

Mathematics
2 answers:
omeli [17]3 years ago
7 0

Answer:

Step-by-step explanation:

Recognize from examining this function h( x ) = ( x + 1 )^2 - 4 that it is a quadratic function whose graph is a parabola that opens up.  The vertex is at (h, k):  (-1, -4).  The axis of symmetry is the vertical line passing through the vertex:  x = -1.

To find the y-intercept, let x = 0 and solve for y:  h(0) = (0 + 1)^2 - 4 = -3.  Thus, the y-intercept is (-3, 0).

To find the x-intercepts, set y = 0 and solve for x:  0 = h(x) = (x + 1)^2 - 4, or

(x + 1)^2 = 4.  Taking the square root of both sides, we get x + 1 = ±2, whose roots are x = 1 and x = -3:  (1, 0) and (-3, 0).

Graph all of these points and the axis of symmetry.  Sketch the parabola that goes through these points.

Vera_Pavlovna [14]3 years ago
3 0

Answer:

h( x ) = ( x + 1 )^2 - 4 x(h)×(÷×1)^2-4

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Match each series with the equivalent series written in sigma notation
PIT_PIT [208]

Answer:

3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n

4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n

2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n

3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n

Step-by-step explanation:

Given

See attachment for complete question

Required

Match equivalent expressions

Solving (a):

3 + 12 + 48 + 192 + 768

The expression can be written as:

3 \to 3*4^{0 --- 0

12 \to 3 * 4^{1 ---- 1

48 \to 3 * 4^{2 --- 2

192 \to 3 * 4^{3 ---- 3

768 \to 3 * 4^{4 ---- 4

For the nth term, the expression is:

Term = 3 * 4^{n ---- n

So, the summation is:

3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n

Solving (b):

4 + 32 + 256 + 2048 + 16384

The expression can be written as:

4 \to 4 * 8^0 --- 0

32 \to 4 * 8^1 ---- 1

256 \to 4 * 8^2 --- 2

2048 \to 4 * 8^3 ---- 3

16384 \to 4 * 8^4 ---- 4

For the nth term, the expression is:

Term \to 4 * 8^n ---- n

So, the summation is:

4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n

Solving (c):

2 + 6 + 18 + 54 + 162

The expression can be written as:

2 \to 2 * 3^0 --- 0

6 \to 2 * 3^1 ---- 1

18 \to 2 * 3^2 --- 2

54 \to 2 * 3^3 ---- 3

162 \to 2 * 3^4 ---- 4

For the nth term, the expression is:

Term \to 2 * 3^n ---- n

So, the summation is:

2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n

Solving (d):

3 + 15 + 75 + 375 + 1875

The expression can be written as:

3 \to 3 * 5^0 --- 0

15 \to 3 * 5^1 ---- 1

75 \to 3 * 5^2 --- 2

375 \to 3 * 5^3 ---- 3

1875 \to 3 * 5^4 ---- 4

For the nth term, the expression is:

Term \to 3 * 5^n ---- n

So, the summation is:

3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n

5 0
2 years ago
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