please help me The graphs of f(x) and g(x) are shown below: graph of function f of x open upward and has its vertex at negative 7, 0. Graph of function g of x opens upward and has its vertex at negative 9, 0. If f(x) = (x + 7)2, which of the following is g(x) based on the translation?
http://broward.flvs.net/webdav/assessment_images/educator_algebraI_v20/segment2_graph33.gif
I'm sorry, I'm not able to see your graphs
@Michele_Laino sorry i had an emergency hold on
@Michele_Laino
hint: the graph of f(x), can be obtained from the graph of g(x), making a traslation by 2 units to right
or, vice versa, the graph of g(x), can be obtained from the graph of f(x), making a traslation by2 units to left
the last condition, can be expressed by this formula: \[\Large g\left( x \right) = f\left( {x + 2} \right)\]
im still sort of confused
If f(x) = (x + 7)2, which of the following is g(x) based on the translation? g(x) = (x + 9)2 g(x) = (x + 5)2 g(x) = (x − 9)2 g(x) = (x − 5)2 theres are the choice they gave me
in other words, you have to replace x, with x+2 into the expression of f(x), namely: \[\Large g\left( x \right) = f\left( {x + 2} \right) = {\left( {x + 2 + 7} \right)^2} = ...\]
o ok it said +5 so i was totally confused
ok so i understand it would be A then?
no, sorry I have made a typo
so it isnt A?
yes! correct! it is option A
ok thank you so much can you help with a few more?
ok!
ok 1 sec please
ok im ready Which graph shows the quadratic function y = 3x2 + 12x + 14?
@Michele_Laino
your function is a parabola, right?
yes it is
the equation of the axis of your parabola, is: \[\Large x = - \frac{{12}}{{2 \cdot 3}} = ...\]
which is the x-coordinate of the vertex. please complete: \[\Large x = - \frac{{12}}{{2 \cdot 3}} = ...?\]
-2?
correct! the equation of the axis of your parabola, is x=-2, furthermore, the x-coordinate of the vertex of your parabola, is also x=-2
so the answer would be graph a because its the only one with negative 2 aas the x coordinate
o nvm it could also be c
but i think it is a
ok! we have to understand what is the right graph: A or C?
i think a because its at -2,-2
the y-coordinate of your parabola is: \[\Large y = \frac{{4ac - {b^2}}}{{4a}}\] where a=3, b=12, and c=14
oops..it is the y-coordinate of the vertex of your parabola
so it is not a?
so after a substitution, we get: \[\Large y = \frac{{4ac - {b^2}}}{{4a}} = \frac{{4 \cdot 3 \cdot 14 - {{12}^2}}}{{4 \cdot 3}} = ...?\] please complete
1sec let me do it
ok!
-2 again so it is a
are you sure? I got a different result
\[\large y = \frac{{4ac - {b^2}}}{{4a}} = \frac{{4 \cdot 3 \cdot 14 - {{12}^2}}}{{4 \cdot 3}} = \frac{{168 - 144}}{{12}} = ...?\]
yes that what i got 168 - 144 i did it wrong its positive 2 but i used thos step
so its c for some reason i added a negtive signat the end of the problem
correct! it is option C
thank you so much
:)
can u help with 1 more
Which of the following represents the factored form of f(x) = x3 − 64x?
@Michele_Laino
ok! I'm here
k can you help me with this 1 Which of the following represents the factored form of f(x) = x3 − 64x?
yes!
first step: we can factor out x, so we can write this: \[\Large {x^3} - 64x = x\left( {{x^2} - 64} \right)\]
subsequently, we can use this algebraic identity: \[\Large {A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)\] where A=x, and B= 8
we have to foil this right if so can you show me how i would set it up?
no, it is not necessary to apply the foil method, since we have to apply that standard identity
you should get this: \[\Large {x^2} - 64 = \left( {x - 8} \right)\left( {x + 8} \right)\]
theses are the choices f(x) = x(x + 8)(x − 8) f(x) = (x − 8)(x + 8) f(x) = x(x − 8)2 f(x) = x(x2 − 8)
so, the complete factorization, is: \[\Large {x^3} - 64x = x\left( {{x^2} - 64} \right) = x\left( {x - 8} \right)\left( {x + 8} \right)\]
o ok so i just needed to find wat squared equals 64 ok so its b
I think it is option A
o ok you are right i see my error with the x variable
ay thank you so much michele u were really helpfull :) lol and u have a sexy as name
thanks! :)
Join our real-time social learning platform and learn together with your friends!