All categories

3 years ago

Which describes the effect of the transformations on the graph of ƒ(x) = x2 when changed to ƒ(x) = −

Mathematics

1 answer:

storchak [24]3 years ago

4 0

Answer:

B) reflected over x-axis, compressed vertically, shifted right 2 units, and shifted up 3 units

Step-by-step explanation:

ƒ(x) = −⅛(x − 2)² + 3

- minus indicates reflection along the x-axis

⅛ indicates a vertical compression

x - 2 indicates 2 units towards right

+3 indicates 3 units upwards

You might be interested in

3e≤6 a e=1 b e=2 c e=3

lesya [120]

A and b are both accurate

For a:
$3(1) \leqslant 6 = 3 \leqslant 6$
which is true

For b:
$3(2) \leqslant 6 = 6 \leqslant 6$
which is also true

7 0

3 years ago

A store has a 28% off sale on hammers. Before the discount, the price of a hammer is $49. What is the sale price?

zheka24 [161]

28% of 49 is 13.72. So subtract that from 49 and you get 35.28. 35.28 is the price after the sale

5 0

2 years ago

Write the algebraic expression that is represented by the model

Salsk061 [2.6K]

3x+2y=5

Hop this helps

4 0

3 years ago

Read 2 more answers

Please help with this work I can't seem to find the answers to any of them, THANK YOU!

Brilliant_brown [7]

Answer:

Dude, that was not nice he just wants to know!

Step-by-step explanation:

6 0

2 years ago

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

Step(i):-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

Step(ii):-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

Step(iii):-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

Final answer:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

3 0

2 years ago