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3 years ago

Which graph shows the axis of symmetry for the function f(x) = (x – 2)2 + 1?

Mathematics

2 answers:

viva [34]3 years ago

8 0

Step-by-step explanation:

Please find the attachment of graph representing the axis of symmetry.

We are asked to choose the graph from our given choices, which shows the axis of symmetry for the function:

$f(x)=(x-2)^2+1$

We can see that our function represents a parabola.

Since we know that axis of symmetry is a vertical line that divides parabola into two congruent parts. The axis of symmetry intersects parabola at its vertex.

Let us find the vertex of our given parabola using formula: $x=\frac{-b}{2a}$ , where,

a = Coefficient of $x^{2}.$

b = Coefficient of x.

To apply this formula we will represent our parabola as a quadratic equation $(ax^2+bx+c)$ as:

$f(x)=x^2-4x+4+1$

$f(x)=x^2-4x+5$

Upon substituting a= 1 and b= -4 in the formula we will get,

$x=\frac{-(-4)}{2*1}$

$x=\frac{4}{2}$

$x=2$

Therefore, the vertical line passing through point x = 2 will be the axis of symmetry for our given function and graph B is the correct choice.

svet-max [94.6K]3 years ago

5 0

(x -2)^2 + 1 = 0
x^2 -4x + 5 = 0
The 'x' value of the axis of symmetry = 'x' value of the vertex.
To get this value, take the derivative of x^2 -4x + 5 = 0 which equals
2x -4 =0
x = 2 which equals the axis of symmetry

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The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest

Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

$P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)$

Calculation the value from standard normal z table, we have,

$P(x < 5) =0.7287= 72.87\%$

b) P( port handles 3 or more million tons of cargo per week)

$P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)$

Calculating the value from the standard normal table we have,

$1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%$

c)P( port handles between 3 million and 4 million tons of cargo per week)

$P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%$

$P(3 \leq x \leq 4) = 23.7\%$

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85$

$= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85$

$=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15$

Calculation the value from standard normal z table, we have,

$P( z \leq -1.036) = 0.15$

$\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65$

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

8 0

3 years ago

HELP ME PLEASE ON THIS

Anvisha [2.4K]

The answer is [ all real numbers ]

The domain are the input values which makes the function defined and real.

Since we don't have any constraints or any undefined points, the domain could be literally anything. Thus proves out answer.

Best of Luck!

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mixer [17]

False it is not it just said that if B is 100 then A is 35

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Temka [501]

Answer:

do you still need this answered?

Step-by-step explanation:

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noname [10]

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