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Algebra 8 Online
OpenStudy (anonymous):

How do I find the vertex of T(x) = (x + 5)3 + 7?

OpenStudy (anonymous):

Well, T(x) = (x + 5)^3 + 7 :)

OpenStudy (anonymous):

i assume it is \[T(x)=(x+3)^2+7\] maybe? because a cubic polynomial does not have a vertex

OpenStudy (anonymous):

No, the question is "In your lab, a substance's temperature has been observed to follow the function T(x) = (x + 5)^3 + 7. The turning point of the graph is where the substance changes from a liquid to a solid. Explain to your fellow scientists how to find the turning point of this function, using complete sentences."

OpenStudy (anonymous):

I assumed that the "turning point" would be the vertex. Am I wrong?

OpenStudy (anonymous):

i guess it "turns" at \((-5,7)\) but that is not a vertex quadratics have vertices

OpenStudy (anonymous):

Is there an equation, or steps to go by to find (-5), 7? I remember learning this, but I would not want to submit it without any work.

OpenStudy (anonymous):

*(-5,7)

OpenStudy (anonymous):

i take it this is not calculus right?

OpenStudy (anonymous):

No, I am in my Junior Year, Algebra 2.

OpenStudy (anonymous):

ok, then it looks like \(f(x)=x^3\) which "turns" at \((0,0)\) but in your case you have \[T(x)=(x+5)^3+7\] which is exactly the same as \(y=x^3\) except shifted left \(5\) units and up \(7\) units

OpenStudy (anonymous):

there is no "equation" or work really, except to look with your eyes and compare \(f(x)\) with \(f(x-h)+k\) and see that the second is identical to the first but shifted left \(h\) units (assuming \(h>0\) and up or down \(k\) units depending on the sign of \(k\)

OpenStudy (anonymous):

Okay, thank you so much. I'll see what I can do. :)

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