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3 years ago

PLEASE HELP!!! MIDTERM!!

Mathematics

2 answers:

Paladinen [302]3 years ago

8 0

Answer:

Step-by-step explanation:

2nd answer choice my guy.... Technically B. :33

dusya [7]3 years ago

7 0

Answer:

Not going to explain my work, because I just did. Somehow just now it dissapeared. That's weird, and frustrating, brainly....

I'll make it simple.

3/20 divided by 3 is 3, now multiply by 20 to get x.

3/20 out of X = 9; X = 60.

There you go.

Brainliest, pl0x?

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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

$x=h$

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

8 0

3 years ago