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Nesterboy [21]
3 years ago
12

Which absolute value function has a graph that is wider than the parent function when graphed represents the parent function f(x

)=|x| reflected over the x-axis and translated 1 unit to the right
A. F(x) = -|x|+1
B. F(x) = -|x-1|
C. F(x) =|-x|+1
D. F(x) =|-x-1|

Mathematics
2 answers:
Advocard [28]3 years ago
7 0
The answer to this is B
pantera1 [17]3 years ago
5 0

Answer:

B.F(x)=-\mid{x-1}\mid

Step-by-step explanation:

We are given that a function

f(x)=\mid x\mid

We have to find absolute value of function when parent function reflected over the x- axis and translated 1 unit to the right.

The transformation rule when the point (x,y) is reflected over x- axis is given by

(x,y)\rightarrow (x,-y)

Apply the reflection on function  over x- axis then, we get

g(x)=-\mid x\mid

Transformation rule when the point (x,y) is shifted a units  towards right is given by

(x,y)\rightarrow (x-a,y)

When the graph is translated 1 units towards right then , obtained function is given by

F(x)=-\mid{x-1}\mid

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5x − 3y = −11<br> 2x − 6y = −14
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In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En
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Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

\frac{dy}{dx} = 3t^{2} - 2t - 27

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

3t^{2} - 2t - 27 = 0

So

a = 3, b = -2, c = -27

Then

\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328

So

t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35

t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55

y = t^{2} - 4 = (3.35)^2 - 4 = 7.22

The first point is (27.55, 7.22)

t = -2.685

x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3

y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21

The second point is (-11.3, 3.21).

8 0
3 years ago
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