Identify the parent function of the equation below.
1 answer:
Answer:
y = x² ⇒ answer C
Step-by-step explanation:
* Lets explain the problem
∵ The function is y = 1/2 x² + 3 , that means the parent function is
transformed by some transformation
- If the parent function is y = x²
- y = x² is a quadratic function represented graphically by upward
parabola with vertex (0 , 0)
∵ y = x² changed to y = 1/2 x² that means we multiply each
y-coordinates of the points on the graph of the function by 1/2
∴ The graph of the function is compressed vertically by scale
factor 1/2
- That means the graph is squeezing toward the x-axis
∵ y = 1/2 x² changed to y = 1/2 x² + 3 that means we add each
y-coordinates of the points on the graph of the function by 3
∴ The function translated 3 units up
- That means the vertex of the parabola changed to (0 , 3)
∴ The new function is y = 1/2 x² + 3
* The parent function of y = 1/2 x² + 3 is y = x²
# Look to the attached graph for more understand
- y = x² ⇒ the red graph (parent function)
- y = 1/2 x² + 3 ⇒ the blue graph
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Yay, easy peasy, nice refresher
53. evaluate for x=infinity
we get 100/(infiniy^2-5) which is basically 100/infinity which is a very small number so about 0
54.
vertical assemtotes, neato
that is where the denomenator equals 0
but wait, simplify first
ok, fraction is simplified (anything factored out would be a hole)
so set the denom to 0 and solve
x^2-5=0
(x-√5)(x+√5)=0
x=-√5 and √5
vertical assemtotes at x=-√5 and x=√5
55. horizontal assemtotes
for the function p(x)/f(x)
if p(x) has lower degree than f(x) then y=0 is the horizontal assemtote
if p(x) has equal degree to f(x) then divide the leading coefients of p(x) and f(x)
if p(x) has greater degree than f(x) then we got slant assemtotes
100/(x^2-5)
100 has lower degree than x^2
y=0 is the horizontal assemtote
56. evalate for x=-5
we get -150/-10=150/10=15
the limit is 15
57. that negative on the right side means that we must approach it from the left
let's see if we can factor out the denomenator
nope
so go and evaluate values like 4, 4.5, 4.9, etc, approaching x=5 from the left
we keep getting smaller and smaller valuse
this make sense because as x approaches 5, we get 30(c)/(c-5) where c is close to 5, and as c appraoches 5 very close, the top part is appraoching 150 but the bottom approaches a very small negative number, and if we do something like 150/-0.00000000001, we get a very negative number
the limit is negative infinity
alternately, you could figure that there is a vertical assemtote at x=5 so we can see that if we aproximaate closer and closer values, we get smaller values. so from what we know about assemtotes, expecially vertical ones, we say that it approaches negative infinity
58.
evalauating we get infintity/inifinty, not helpful
hmmm, is there a horizontal assemtote to help us?
the degree of top is higher so we got a slant assemtote
the only way is to use long division to divide the top by the bottom (see attachment)
we get
the slant assemtote is
as you can see, as x approaches infinity, y also will approach infinity
ANSWERS:
53. 0
54. VA is x=-√5 and x=√5
55. HA is y=0
56. the limit is 15
57. the limit is -∞
58. the limit is ∞
53. evaluate for x=infinity
we get 100/(infiniy^2-5) which is basically 100/infinity which is a very small number so about 0
54.
vertical assemtotes, neato
that is where the denomenator equals 0
but wait, simplify first
ok, fraction is simplified (anything factored out would be a hole)
so set the denom to 0 and solve
x^2-5=0
(x-√5)(x+√5)=0
x=-√5 and √5
vertical assemtotes at x=-√5 and x=√5
55. horizontal assemtotes
for the function p(x)/f(x)
if p(x) has lower degree than f(x) then y=0 is the horizontal assemtote
if p(x) has equal degree to f(x) then divide the leading coefients of p(x) and f(x)
if p(x) has greater degree than f(x) then we got slant assemtotes
100/(x^2-5)
100 has lower degree than x^2
y=0 is the horizontal assemtote
56. evalate for x=-5
we get -150/-10=150/10=15
the limit is 15
57. that negative on the right side means that we must approach it from the left
let's see if we can factor out the denomenator
nope
so go and evaluate values like 4, 4.5, 4.9, etc, approaching x=5 from the left
we keep getting smaller and smaller valuse
this make sense because as x approaches 5, we get 30(c)/(c-5) where c is close to 5, and as c appraoches 5 very close, the top part is appraoching 150 but the bottom approaches a very small negative number, and if we do something like 150/-0.00000000001, we get a very negative number
the limit is negative infinity
alternately, you could figure that there is a vertical assemtote at x=5 so we can see that if we aproximaate closer and closer values, we get smaller values. so from what we know about assemtotes, expecially vertical ones, we say that it approaches negative infinity
58.
evalauating we get infintity/inifinty, not helpful
hmmm, is there a horizontal assemtote to help us?
the degree of top is higher so we got a slant assemtote
the only way is to use long division to divide the top by the bottom (see attachment)
we get
the slant assemtote is
as you can see, as x approaches infinity, y also will approach infinity
ANSWERS:
53. 0
54. VA is x=-√5 and x=√5
55. HA is y=0
56. the limit is 15
57. the limit is -∞
58. the limit is ∞