SOLUTION: Hi, again thanks for all the help it is great during finals week.
Find the vertex and intercepts then graph the equation state if the concave is up or down.
f(x)=3(x-2)^2-3.
Tha
Algebra ->
Quadratic-relations-and-conic-sections
-> SOLUTION: Hi, again thanks for all the help it is great during finals week.
Find the vertex and intercepts then graph the equation state if the concave is up or down.
f(x)=3(x-2)^2-3.
Tha
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Question 144378: Hi, again thanks for all the help it is great during finals week.
Find the vertex and intercepts then graph the equation state if the concave is up or down.
f(x)=3(x-2)^2-3.
Thanks. Answer by nabla(475) (Show Source):
You can put this solution on YOUR website! f(x)=3(x-2)^2-3
Notice that (x-2)^2 is always going to be >=0. Thus, the absolute minimum (we know it is a minimum due to the reasoning supplied for concavity below) of this function will be where that expression is 0. Namely, f(2)=-3. By definition this will be the vertex. Now, the function must open upward because the x^2 term will be positive with a coefficient of 3 upon expansion. x^2 is the driving term in this function.
Since it opens upward, we will have upward concavity.
As for intercepts:
x-ints: 0=3(x-2)^2-3
is the same as 0=(x-2)^2-1
is the same as 0=x^2-4x+3
is the same as 0=(x-3)(x-1)
which implies x=3 or x=1.
f(3)=0. This gives (3,0)
f(1)=0. This gives (1,0).
y-int: y=3(0-2)^2-3=9. This gives (0,9).
I will graph this even though it's not asked for:
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