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muminat
2 years ago
14

Drag each tile to the correct box. Graph the functions as transformations of f(x)=x^2. Arrange the parabolas with respect to the

position of their vertices from left to right. 1: y=f(x)
2: y=f(x-4)
3: y=f(x+3)
4: y=f(x-7)
5: y=f(x+2)
6:y=f(x-1/2)
Mathematics
1 answer:
rusak2 [61]2 years ago
8 0

If a constant is added to a variable then the graph gets shifted toward left and vice-versa. Then the graphs are shown below.

<h3>What is the parabola?</h3>

It's the locus of a moving point that keeps the same distance between a stationary point and a specified line. The focus is a non-movable point, while the directrix is a non-movable line.

The parabola is given below.

f(x) = x²

Then the parabolas with respect to the position of their vertices from left to right will be

1: y = f(x) = x²

2: y = f(x - 4) = (x - 4)²

3: y = f(x + 3) = (x + 3)²

4: y = f(x - 7) = (x - 7)²

5: y = f(x + 2) = (x + 2)²

6: y = f(x - 1/2) = (x - 1/2)²

The graph of the parabola is given below.

More about the parabola link is given below.

brainly.com/question/8495504

#SPJ1

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8 0
4 years ago
Read 2 more answers
Let e1= 1 0 and e2= 0 1 ​, y1= 4 5 ​, and y2= −2 7 ​, and let​ T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps
Furkat [3]

Answer:

The image of \left[\begin{array}{c}4&-4\end{array}\right] through T is \left[\begin{array}{c}24&-8\end{array}\right]

Step-by-step explanation:

We know that T: IR^{2}  → IR^{2} is a linear transformation that maps e_{1} into y_{1} ⇒

T(e_{1})=y_{1}

And also maps e_{2} into y_{2}  ⇒

T(e_{2})=y_{2}

We need to find the image of the vector \left[\begin{array}{c}4&-4\end{array}\right]

We know that exists a matrix A from IR^{2x2} (because of how T was defined) such that :

T(x)=Ax for all x ∈ IR^{2}

We can find the matrix A by applying T to a base of the domain (IR^{2}).

Notice that we have that data :

B_{IR^{2}}= {e_{1},e_{2}}

Being B_{IR^{2}} the cannonic base of IR^{2}

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]   (Data of the problem)

T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]   (Data of the problem)

Writing the matrix A :

A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]

Now with the matrix A we can find the image of \left[\begin{array}{c}4&-4\\\end{array}\right] such as :

T(x)=Ax ⇒

T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]

We found out that the image of \left[\begin{array}{c}4&-4\end{array}\right] through T is the vector \left[\begin{array}{c}24&-8\end{array}\right]

3 0
3 years ago
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