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y-x-2-2-3-Translation-T-1-a-b-y-x-2-a-b-




Question Number 8345 by suci last updated on 09/Oct/16
y=(x+2)^2 −3  Translation T_1 = ((a),(b) )  y′=x^2   a=? b=?
$${y}=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3} \\ $$$${Translation}\:{T}_{\mathrm{1}} =\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix} \\ $$$${y}'={x}^{\mathrm{2}} \\ $$$${a}=?\:{b}=? \\ $$
Answered by sandy_suhendra last updated on 09/Oct/16
y=(x+2)^2 −3   has extreme point at (−2 , 3)  y=x^2   has extreme point at (0 , 0)  so   a=2  and  b=−3
$$\mathrm{y}=\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}\:\:\:\mathrm{has}\:\mathrm{extreme}\:\mathrm{point}\:\mathrm{at}\:\left(−\mathrm{2}\:,\:\mathrm{3}\right) \\ $$$$\mathrm{y}=\mathrm{x}^{\mathrm{2}} \:\:\mathrm{has}\:\mathrm{extreme}\:\mathrm{point}\:\mathrm{at}\:\left(\mathrm{0}\:,\:\mathrm{0}\right) \\ $$$$\mathrm{so}\:\:\:\mathrm{a}=\mathrm{2}\:\:\mathrm{and}\:\:\mathrm{b}=−\mathrm{3} \\ $$

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