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3 years ago

13

What is the value of f (x)=16^x when x=1/2 ? A. 2 B. 4 C. 8 D. 32

1 answer:

Cerrena [4.2K]3 years ago

8 0

Answer:

B

Step-by-step explanation:

Using the rule of exponents

$a^{\frac{m}{n} }$ ⇔ $\sqrt[n]{a^{m} }$

Given

f(x) = $16^{x}$ , then when x = $\frac{1}{2}$

f( $\frac{1}{2}$ ) = $16^{\frac{1}{2} }$ = $\sqrt[2]{16}$ = 4

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The graph of this function is shifted downwards and the axis of symmetry remains x=1. Which function below the equation of the n

Natali5045456 [20]

Answer:

$y=-x^{2}+2x$

$y=-x^{2}+2x-4$

$y=-x^{2}+2x-3$

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

The axis of symmetry is equal to the x-coordinate of the vertex

so

$x=h$

If a> 0 then the parabola open upward (vertex is a minimum)

If a< 0 then the parabola open downward (vertex is a maximum)

In this problem we have

$y=-x^{2} +2x+3$

The vertex is the point $(1,4)$ ------> observing the graph

The axis of symmetry is $x=1$

If the graph of this function is shifted downwards and the axis of symmetry remains x=1

then

The x-coordinate of the vertex of the new graph must be equal to 1

The y-coordinate of the vertex of the new graph must be less than 4

The parabola of the new graph open downward

therefore

<u>Verify each case</u>

case a) $y=-x^{2}+2x$

Convert to vertex form

$y=-(x^{2}-2x)$

$y-1=-(x^{2}-2x+1)$

$y-1=-(x-1)^{2}$

$y=-(x-1)^{2}+1$

The vertex is (1,1)

therefore

The function could be the equation of the new graph

case b) $y=-x^{2}-2x+3$

Convert to vertex form

$y-3=-(x^{2}+2x)$

$y-3-1=-(x^{2}+2x+1)$

$y-4=-(x+1)^{2}$

$y=-(x+1)^{2}+4$

The vertex is (-1,4)

therefore

The function cannot be the equation of the new graph

case c) $y=-x^{2}+2x-4$

Convert to vertex form

$y+4=-(x^{2}-2x)$

$y+4-1=-(x^{2}-2x+1)$

$y+3=-(x-1)^{2}$

$y=-(x-1)^{2}-3$

The vertex is (1,-3)

therefore

The function could be the equation of the new graph

case d) $y=-x^{2}+2x+4$

Convert to vertex form

$y-4=-(x^{2}-2x)$

$y-4-1=-(x^{2}-2x+1)$

$y-5=-(x-1)^{2}$

$y=-(x-1)^{2}+5$

The vertex is (1,5)

therefore

The function cannot be the equation of the new graph

case e) $y=-x^{2}+2x-3$

Convert to vertex form

$y+3=-(x^{2}-2x)$

$y+3-1=-(x^{2}-2x+1)$

$y+2=-(x-1)^{2}$

$y=-(x-1)^{2}-2$

The vertex is (1,-2)

therefore

The function could be the equation of the new graph

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Leto [7]

Answer:

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Step-by-step explanation:

and u should've paid attention during class

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