Let T:R²->R² be a linear transformation ,and assume that T (1,2)=(-1,1) and T(1,-1)=(2,3)
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Answer:
12. Range = -∞ < x <+∞; Domain = y > 0
13. Range = -∞ < x <+∞; Domain = y > 3
Step-by-step explanation:
12. f(x) = ½(6)^x
(a) Create a table containing a few values of x and y
x y
-2 ½(6)⁻² = ½(⅙)² = ½ × ¹/₃₆ = ¹/₇₂ = 0.01
-1 ½(6)⁻¹ = ½(⅙) = ½ × ¹/₁₂ = ¹/₂₄ = 0.08
0 ½(6)⁰ = ½ × 0 = ½ = 0.5
1 ½(6)¹ = ½ × 6 = 3
2 ½(6)² = ½ × 36 = 18
(b) Draw your axes
Let the x-axis run from -5 to +5 and the y-axis from 0 to 20
(c) Plot your points
They should look like Fig. 1 below.
(d) Draw the graph
Draw a smooth line through the points.
Extend the line in both directions to the edges of the graph.
Your graph should look like Fig. 2.
(e) Identify the domain and range of the function
The <em>domain</em> is the set of all possible x-values that will give real values for y.
It looks like x can vary from -∞ to +∞ and give a real value for y.
The <em>domain is -∞ < x <+∞</em>.
The <em>range</em> of a function is the spread of all possible y-values.
It looks like y ⟶ ∞ as x ⟶ ∞, but y ⟶ 0 as x ⟶ -∞.
The x-axis is an <em>asymptote</em>. The function can get as close as possible to the x-axis, but it can never be zero or negative.
The range is y > 0.
12. f(x) = 2^x +3
(a) Create a table containing a few values of x and y
<u> x </u> <u> y </u>
-4 2⁻⁴ + 3 = (½)⁴ + 3 = ¹/₁₆ + 3 = 3¹/₁₆ = 3.1
-2 2⁻² + 3 = (½)² + 3 = ¼ + 3 = 3¼ = 3.2
0 2⁰ + 3 = 1 + 3 = 4
2 2² + 3 = 4 + 3 = 7
4 2⁴ + 3 = 16 + 3 = 19
(b) Draw your axes
(c) Plot your points
(d) Draw the graph
Your graph should look like Fig. 3.
(e) Identify the domain and range of the function
The domain is -∞ < x <+∞.
It looks like y ⟶ ∞ as x ⟶ ∞, but y ⟶ 3 as x ⟶ -∞. The line y = 3 is an asymptote.
The range is y > 3.
