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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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olchik [2.2K]

Answer:

f(-4) = -2

Step-by-step explanation:

To find f(-4) use the "piece" of the function rule defined for x < -3 (because -4 is less than -3).  That part of the rule says the output (range) value is x + 2, so f(-4) = -4 + 2 = -2.

5 0
2 years ago
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Oksana_A [137]
The slope is 4/3 the unit rate is 1
3 0
3 years ago
Solve for x: 3 − (2x − 5)&lt; −4(x + 2
guapka [62]

Answer:x<−8

Step-by-step explanation:

8 0
2 years ago
What is the slope of the line passing through (16, -2) and (-30, 6)?
Novay_Z [31]

m=\frac{\Delta y}{\Delta x}=\frac{-2-6}{16-(-30)}=\frac{-8}{46}=\boxed{-\frac{4}{23}}.

Hope this helps.

5 0
2 years ago
Read 2 more answers
P(x) = x + 1x² – 34x + 343<br> d(x)= x + 9
Feliz [49]

Answer:

x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > \frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1. Go ahead and simplify.

x=\frac{9}{d-1};\quad \:d\ne \:1.

Now, \mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}.

P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343.

Group the like terms... 1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343.

\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1} > 1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343.

Now for 1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}.

\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}.

Now for 33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}.

Thus we then get \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343.

Now we want to combine fractions. \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}.

\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2

\mathrm{For}\:\frac{297}{d-1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:d-1 > \frac{297}{d-1}=\frac{297\left(d-1\right)}{\left(d-1\right)\left(d-1\right)}=\frac{297\left(d-1\right)}{\left(d-1\right)^2}

\frac{9^2}{\left(d-1\right)^2}-\frac{297\left(d-1\right)}{\left(d-1\right)^2} > \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}> \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}.

Expand 81-297\left(d-1\right) > -297\left(d-1\right) > \mathrm{Apply\:the\:distributive\:law}: \:a\left(b-c\right)=ab-ac.

-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1.

\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}

Therefore P=\frac{-297d+378}{\left(d-1\right)^2}+343.

Hope this helps!

5 0
3 years ago
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