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Stella [2.4K]
4 years ago
10

How to solve piecewise functions step by step?

Mathematics
1 answer:
Zanzabum4 years ago
5 0
If you can graph a function, you can graph piecewise functions.  Each one of them is a different line or curve within the domain for that specific line of curve.  If the domain states it's less than or greater than a number, you circle that point on the line.  If the domain states it's also possibly equal to the point at the beginning or the end, you make a closed dot.  

I"m sure that's clear as mud, please respond with any additional questions.
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AlexFokin [52]
I have no idea i’m sorry
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A fleet of vehicles is comprised of 60 vans, 20 limos, and X sedans. If 10% of all vehicles are limos, how many sedans are in th
xz_007 [3.2K]
10% of all the cars is 20 limos.

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x = 120 sedans
4 0
4 years ago
Read 2 more answers
Can someone please help me with this? i still have 2 more pages to do and I'm stressed out of my mind I honestly just wanna pass
melisa1 [442]
1. First we are going to find the vertex of the quadratic function f(x)=2x^2+8x+1. To do it, we are going to use the vertex formula. For a quadratic function of the form f(x)=ax^2+bx +c, its vertex (h,k) is given by the formula h= \frac{-b}{2a}; k=f(h).

We can infer from our problem that a=2 and b=8, sol lets replace the values in our formula:
h= \frac{-8}{2(2)}
h= \frac{-8}{4}
h=-2

Now, to find k, we are going to evaluate the function at h. In other words, we are going to replace x with -2 in the function:
k=f(-2)=2(-2)^2+8(-2)+1
k=f(-2)=2(4)-16+1
k=f(-2)=8-16+1
k=f(-2)=-7
k=-7
So, our first point, the vertex (h,k) of the parabola, is the point (-2,-7).

To find our second point, we are going to find the y-intercept of the parabola. To do it we are going to evaluate the function at zero; in other words, we are going to replace x with 0:
f(x)=2x^2+8x+1
f(0)=2(0)^2+(0)x+1
f(0)=1
So, our second point, the y-intercept of the parabola, is the point (0,1)

We can conclude that using the vertex (-2,-7) and a second point we can graph f(x)=2x^2+8x+1 as shown in picture 1.

2. The vertex form of a quadratic function is given by the formula: f(x)=a(x-h)^2+k
where
(h,k) is the vertex of the parabola.

We know from our previous point how to find the vertex of a parabola. h= \frac{-b}{2a} and k=f(h), so lets find the vertex of the parabola f(x)=x^2+6x+13.
a=1
b=6
h= \frac{-6}{2(1)}
h=-3
k=f(-3)=(-3)^2+6(-3)+13
k=4

Now we can use our formula to convert the quadratic function to vertex form:
f(x)=a(x-h)^2+k
f(x)=1(x-(-3))^2+4
f(x)=(x+3)^2+4

We can conclude that the vertex form of the quadratic function is f(x)=(x+3)^2+4.

3. Remember that the x-intercepts of a quadratic function are the zeros of the function. To find the zeros of a quadratic function, we just need to set the function equal to zero (replace f(x) with zero) and solve for x.
f(x)=x^2+4x-60
0=x^2+4x-60
x^2+4x-60=0
To solve for x, we need to factor our quadratic first. To do it, we are going to find two numbers that not only add up to be equal 4 but also multiply to be equal -60; those numbers are -6 and 10.
(x-6)(x+10)=0
Now, to find the zeros, we just need to set each factor equal to zero and solve for x.
x-6=0 and x+10=0
x=6 and x=-10

We can conclude that the x-intercepts of the quadratic function f(x)=x^2+4x-60 are the points (0,6) and (0,-10).

4. To solve this, we are going to use function transformations and/or a graphic utility.
Function transformations.
- Translations:
We can move the graph of the function up or down by adding a constant c to the y-value. If c\ \textgreater \ 0, the graph moves up; if c\ \textless \ 0, the graph moves down.

- We can move the graph of the function left or right by adding a constant c to the x-value. If c\ \textgreater \ 0, the graph moves left; if c\ \textless \ 0, the graph moves right.

- Stretch and compression:
We can stretch or compress in the y-direction by multiplying the function by a constant c. If c\ \textgreater \ 1, we compress the graph of the function in the y-direction; if 0\ \textless \ c\ \textless \ 1, we stretch the graph of the function in the y-direction.

We can stretch or compress in the x-direction by multiplying x by a constant c. If c\ \textgreater \ 1, we compress the graph of the function in the x-direction; if 0\ \textless \ c\ \textless \ 1, we stretch the graph of the function in the x-direction.

a. The c value of f(x) is 2; the c value of g(x) is -3. Since c is added to the whole function (y-value), we have an up/down translation. To find the translation we are going to ask ourselves how much should we subtract to 2 to get -3?
c+2=-3
c=-5

Since c\ \textless \ 0, we can conclude that the correct answer is: It is translated down 5 units.

b. Using a graphing utility to plot both functions (picture 2), we realize that g(x) is 1 unit to the left of f(x)

We can conclude that the correct answer is: It is translated left 1 unit.

c. Here we have that g(x) is f(x) multiplied by the constant term 2. Remember that We can stretch or compress in the y-direction (vertically) by multiplying the function by a constant c.

Since c\ \textgreater \ 0, we can conclude that the correct answer is: It is stretched vertically by a factor of 2.

4 0
3 years ago
What is the relationship between the graphs of y = 2x and y = 2-x?
melomori [17]

Answer:

The relationship between the graphs is the intersection point at (0.667,1.333)

Step-by-step explanation:

we have

y=2x ----> equation A

The slope of the line A is equal to m=2

The line passes through the origin

y=2-x ----> equation B

The slope of the line B is m=-1

The x-intercept is the point (2,0)

The y-intercept is the point (0,2)

Line A and Line B are not parallel (the slopes are not equal)

Line A and Line B are not perpendicular (the product of their slopes is not equal to -1)

so

The relationship between Line A and Line B is the intersection point both graphs

using a graphing tool

The intersection point is (0.667,1.333)

see the attached figure

The intersection point is a common point , therefore belongs to both lines

8 0
4 years ago
show that if the diagonals of a quadrilateral are equal and bisect each other at right angles then it is a square​
zvonat [6]

Answer:

Given,

Diagonals are equal

AC=BD                  .......(1)

and the diagonals bisect each other at right angles

OA=OC;OB=OD           ...... (2)

∠AOB= ∠BOC= ∠COD=  ∠AOD= 90  

0

    ..........(3)

 

Proof:

Consider △AOB and △COB

OA=OC  ....[from (2)]

∠AOB= ∠COB

OB is the common side

Therefore,

△AOB≅ △COB

From SAS criteria, AB=CB

Similarly, we prove

△AOB≅ △DOA, so AB=AD

△BOC≅ △COD, so CB=DC

So, AB=AD=CB=DC               ....(4)

So, in quadrilateral ABCD, both pairs of opposite sides are equal, hence ABCD is  parallelogram

In △ABC and △DCB

AC=BD            ...(from (1))

AB=DC            ...(from $$(4)$$)

BC is the common side

△ABC≅ △DCB

So, from SSS criteria, ∠ABC= ∠DCB

Now,

AB∥CD,BC is the tansversal

∠B+∠C= 180  

0

 

∠B+∠B= 180  

0

 

∠B= 90  

0

 

Hence, ABCD is a parallelogram with all sides equal and one angle is 90  

0

 

So, ABCD is a square.

Hence proved.

Step-by-step explanation:

4 0
3 years ago
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