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forsale [732]
3 years ago
8

Graph question, is it correct please

Mathematics
2 answers:
Molodets [167]3 years ago
7 0
That is incorrect, the correct answer is b, i think

LenKa [72]3 years ago
4 0

you are right.

they all go up into a positive region and out of a negative region

hopes this helps 

You might be interested in
Directions: For the following systems, determine whether or not the ordered triple (1, -1, -2) is a solution to the equations be
Fofino [41]

Answer:

<h2>NOT</h2>

Step-by-step explanation:

Put x = 1, y = -1 and z = -2 to the equation 3x - 3y - z = 2, and check the equality:

3(1) - 3(-1) - (-2) = 3 + 3 + 2 = 8 ≠ 2

8 0
3 years ago
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
Let f be the function defined by f(x) = 3x^5-5x^3+2
sveticcg [70]
(a) When f is increasing the derivative of f is positive.

    f'(x) = 15x^4 - 15x^2 > 0
            15x^2(x^2 - 1)> 0
               x^2 - 1    > 0 (The inequality doesn't flip sign since x^2 is positive)
               x^2 > 1
    Then f is increasing when x < -1 and x > 1.

(b) The f is concave upward when f''(x) > 0.
    f''(x) = 60x^3 - 30x > 0
           30x(2x^2 - 1) > 0
             x(2x^2 - 1) > 0
            x(x^2 - 1/2) > 0
x(x - 1/sqrt(2))(x + 1/sqrt(2)) > 0

    There are four regions here. We will check if f''(x) > 0.
    x < -1/sqrt(2):     f''(-1) = -30 < 0
    -1/sqrt(2) < x < 0: f''(-0.5) = 7.5 > 0
    0 < x < 1/sqrt(2):  f''(0.5) = -7.5 < 0
    x > 1/sqrt(2):      f''(1) = 30 > 0

    Thus, f''(x) > 0 at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
    Therefore, f is concave upward at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).

(c) The horizontal tangents of f are at the points where f'(x) = 0
    15x^2(x^2 - 1) = 0
    x^2 = 1
    x = -1 or x = 1
    f(-1) = 3(-1)^5 - 5(-1)^3 + 2 = 4
    f(1) = 3(1)^5 - 5(1)^3 + 2 = 0

    Therefore, the tangent lines are y = 4 and y = 0.
8 0
4 years ago
Read 2 more answers
- Solve the equation. (1 point)
Arada [10]

Answer:

3(x-7) – x=2x-21 <em>( the equation)</em>

--> infinitely many solutions

Step-by-step explanation:

8 0
3 years ago
.................................
enyata [817]

Answer:

0

Step-by-step explanation:

Given:

(-3, 2)

(4, 2)

Required:

Slope that passes through the two points given.

Solution:

Slope = \frac{y_2 - y_1}{x_2 - x_1}

Let,

(-3, 2) = (x_1, y_1)

(4, 2) = (x_2, y_2)

Plug in the values into the formula

Slope = \frac{2 - 2}{4 -(-3)}

= \frac{0}{7}

slope = 0

8 0
3 years ago
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