May someone please help me out with this problem? May you also explain it?
1 answer:
Answer:
Step-by-step explanation:
Given function is,
f(x) =
If the given function is vertically stretched by a scale factor of Or 1.5,
Transformed function will be,
h(x) =
h(x) =
Further function 'h' is shifted 1 units upwards,
g(x) = h(x) + 1
g(x) =
Domain of the function → x ≥ 0 Or [0, ∞)
Range of the function → y ≥ 1 Or [1, ∞)
Transformations done → Parent function f(x) is vertically stretched by a scale factor of then shifted 1 unit upward.
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Answer:
C:
Step-by-step explanation:
The y-axis is the asymptote, and G(x) approaches it as x is getting smaller and smaller.
Actually, even more accurate would be to write
The down arrow shows that x has to approach 0 from the positive side.
Answer:
Step-by-step explanation:
We have to see how the canonical vectors are transformed throught T. Lets first define T in any basis.
Since T is a reflection, then any element of the line y = x/2 if fixed by T. Therefore T(2,1) = (2,1).
On the other hand, any vector perpendicular to the line direction should be sent to its opposite value. We can take, for example, (-1,2) (note that the scalar product (2,1) * (-1,2) = -2+2 = 0). As a consecuence T(-1,2) = (1,-2). We have
- T(2,1) = (2,1)
- T(-1,2) = (1,-2)
By summing the first vector with the double of the second one we get, using linearity
T(0,5) = T( (2,1) + 2(-1,2)) = T(2,1) + 2T(-1,2) = (2,1) + 2(1,-2) = (4,-3)
Hence, T(0,1) = (4/5,-3/5)
Now, we take the second vector and substract it the double of the first one (to kill the second variable)
T(-3,0) = T( (-1,2) - 2*(2,1) ) = T(-1,2) -2T(2,1) = (1,-2)-2(2,1) = (-3,-4)
Therefore, T(1,0) = (1,4/3)
The matrix A induced by T has in its first column T(1,0) and in its second column T(0,1). We conclude that