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Zanzabum
3 years ago
10

In your lab, a substance's temperature has been observed to follow the function T(x) = (x − 1)3 + 5. The turning point of the gr

aph is where the substance changes from a liquid to a gas. Using complete sentences in your written answer, explain to your fellow scientists how to find the turning point of this function. Hint: The turning point of the graph is similar to the vertex of a quadratic function. (10 points)
Mathematics
1 answer:
Iteru [2.4K]3 years ago
4 0
If it is a calculus question then you can find the derivative dT/dx, set it to zero and solve for x.  If this is a pre-calculus question, then you can answer is it as follows: 

The function T = x^3 has a turning point at x = 0. We can determine this simply by looking at the graph T = x^3. Therefore, (0,0) is the turning point for the graph T = x^3. The function T(x) = (x - 4)^3 + 6 will be the graph T = x^3 shifted to the right by 4 units and shifted up by 6 units. Therefore, the turning point will also shift from (0,0) to 4 units to the right and 6 units up. The turning point will be (4,6) for the function T(x) = (x - 4)^3 + 6.

You're welcome :)


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3 years ago
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keeping in mind that the vertex is between the focus point and the directrix, in this cases we have the focus point above the directrix, meaning is a vertical parabola opening upwards, Check the picture below, which means the "x" is the squared variable.

now, the vertical distance from the focus point to the directrix is \bf \cfrac{17}{8}-\cfrac{15}{8}\implies \cfrac{2}{8} , which means the distance "p" is half that or 1/8, and is positive since it's opening upwards.

since the vertex is 1/8 above the directrix, that puts the vertex at \bf \cfrac{15}{8}+\stackrel{p}{\cfrac{1}{8}}\implies \cfrac{16}{8}\implies 2 , meaning the y-coordinate for the vertex is 2.

\bf \textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill

\bf \begin{cases} h=-4\\ k=2\\ p=\frac{1}{8} \end{cases}\implies 4\left(\frac{1}{8} \right)(y-2)=[x-(-4)]^2\implies \cfrac{1}{2}(y-2)=(x+4)^2 \\\\\\ y-2=2(x+4)^2\implies \blacktriangleright y = 2(x+4)^2+2 \blacktriangleleft

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hope this helps you

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