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Kipish [7]
3 years ago
6

Use the drop-down menus to describe the key aspects of the function f(x) = –x2 – 2x – 1. The vertex is the . The function is inc

reasing . The function is decreasing . The domain of the function is . The range of the function is
Mathematics
2 answers:
mixer [17]3 years ago
7 0
Let's dissect    <span>f(x) = –x2 – 2x – 1.  

a) This is a quadratic equation whose graph is a parabola that opens down.  
b) The coefficients of x^2, x and 1 are {-1,-2, -1}.


</span>                                                                             
c) Applying the quadratic formula results in

        -(-2) plus or minus sqrt ([-2]^2 - 4(-1)(-1) )
x = ----------------------------------------------------------
                            2(-1)

or x  = + 2 plus or minus sqrt(0)
           --------------------------------- = -1
                        -2       

Actually, we have a double root here:  x is {-1, -1}.

This means that the graph of this parabola touches the x-axis once, at (-1,0), but does not cross that axis.  (-1,0) is also the "root" or "solution" or "x-intercept"

The function is increasing from -infinity up to -1, and decreasing from -1 to positive infinity.

The domain is "the set of all real numbers."

The range is "the set of all real numbers from neg. infinity to zero."  y is never positive.

Please graph this function, so that you can verify this information.
Ierofanga [76]3 years ago
5 0

Answer:

The vertex is (-1, 0),

The function is increasing and decreasing both for different intervals,

The domain of the function is set of all real numbers,

Range is (-∞, 0]

Step-by-step explanation:

Given function,

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

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

f(x) = -(x+1)^2+0

Since, vertex form of a quadratic equation is,

f(x) = a(x-h)^2 + k

Where,

(h, k) is the vertex of the function,

By comparing,

Vertex of the given function = (-1, 0),

A quadratic function with negative leading coefficient is increasing from -∞ to its vertex and decreasing from its vertex to +∞,

Also, domain of a quadratic function is set of all real numbers,

While, quadratic function with negative leading coefficient is maximum at its vertex,

∴ Range = (-∞, 0]

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Answer:

see the procedure

Step-by-step explanation:

we have

12\ \frac{m}{s}

Remember that

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To convert sec to min multiply by (1/60)

substitute

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Step-by-step explanation:

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Line m and point P are shown below. Part A: Using a compass and straightedge, construct line n parallel to line m and passing th
Arturiano [62]

Answer:

  see the attached

Step-by-step explanation:

In the attachment, we will refer to the circles, bottom-to-top, as circles 1, 2 and 3. The black points of intersection, bottom-to-top, will be referred to by the letters A, B, C, D. The transversal line through the white point (W) and pink point (P) will be line q.

Step 1. Draw line q through point P so it intersects line m at some convenient point. Label that point W.

Step 2. Choose an arbitrary radius for your compass. Here, we have chosen it to be the length WB. It happens to be less than half the length of WP, but that is not a requirement.

Step 3. Draw an arc of the chosen radius centered at W and intersecting line q and line m. Label the intersection points A (on line m) and B (on line q). These intersection points are on circle 1.

Step 4. Draw an arc of the same radius centered at P. It should be a long enough arc that it would intersect the proposed line parallel to m. Label the intersection point on line q with label C. This intersection point is on circle 3.

Step 5. Adjust the compass width (radius) to the distance from A to B. This is the radius of circle 2.

Step 6. Draw an arc centered at C so that it intersects the arc of Step 4. This is circle 2, and you want it to intersect circle 3. Label that point of intersection D.

Step 7. Draw line PD parallel to m.

_____

The point of the construction is to create congruent alternate interior angles AWB and CPD, so that lines AW and PD are parallel.

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