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Question Number 123302 by peter frank last updated on 24/Nov/20

Answered by MJS_new last updated on 24/Nov/20

$$y=a(x-2)(x+2)={ax}^2-4a$$$$3=-3a\Rightarrow a=-1$$$$y=4-x^2$$$$\mathrm{the}\mathrm{rest}\mathrm{is}\mathrm{easy}$$$$$$

Commented by peter frank last updated on 24/Nov/20

$$\mathrm{why}\mathrm{you}\mathrm{exclude}(1,3)?$$

Commented by MJS_new last updated on 24/Nov/20

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}$$$$y=a(x-2)(x+2)\mathrm{with}x=1\mathrm{and}y=3\mathrm{gives}$$$$3=a(-1)\left(3\right)=-3a\Rightarrow a=-1$$

Commented by peter frank last updated on 24/Nov/20

$$\mathrm{sorry}\mathrm{sir}\mathrm{explain}\mathrm{first}\mathrm{line}$$

Commented by MJS_new last updated on 24/Nov/20

$$\mathrm{if}\mathrm{it}\mathrm{passes}\mathrm{through}\left(\begin{array}{c}2\\ 0\end{array}\right)\mathrm{it}\mathrm{also}\mathrm{passes}\mathrm{through}$$$$\left(\begin{array}{c}-2\\ 0\end{array}\right)\mathrm{due}\mathrm{to}\mathrm{symmetry}\mathrm{about}\mathrm{the}y-\mathrm{axis}$$$$\Rightarrow \mathrm{we}\mathrm{know}\mathrm{both}\mathrm{zeros}\Rightarrow y=a(x-2)(x+2)$$

Answered by MJS_new last updated on 24/Nov/20

$$\mathrm{you}\mathrm{can}\mathrm{do}\mathrm{it}\mathrm{like}\mathrm{this}:$$$$y={ax}^2+bx+c$$$$\left(1\right)\left(\begin{array}{c}2\\ 0\end{array}\right)\in \mathrm{par}\Rightarrow 0=4a+2b+c$$$$\left(2\right)\left(\begin{array}{c}1\\ 3\end{array}\right)\in \mathrm{par}\Rightarrow 3=a+b+c$$$$\left(3\right)\mathrm{symmetry}\Rightarrow \left(\begin{array}{c}-2\\ 0\end{array}\right)\in \mathrm{par}\vee \left(\begin{array}{c}-1\\ 3\end{array}\right)\in \mathrm{par}$$$$\Rightarrow 0=4a-2b+c\vee 3=a-b+c$$$$\mathrm{choose}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{lasr}2\mathrm{options}\mathrm{and}\mathrm{solve}$$$$\mathrm{the}\mathrm{system}\mathrm{for}a,b,c$$$$\Rightarrow \mathrm{you}\mathrm{get}a=-1b=0c=4$$$$y=-x^2+4$$

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