Question-74802

Question Number 74802 by liki last updated on 30/Nov/19

Commented by liki last updated on 30/Nov/19

$$…{plz}\:{i}\:{need}\:{help}\:;{mr}\:{w}\:{and}\:{mr}\:{mind}\:{is}\:{power} \\ $$$$,\:{or}\:{anyeone}\:{to}\:{assist}\:{me}! \\ $$

Commented by abdomathmax last updated on 30/Nov/19

we have f(x)=4x^2 −4x+4 the equation of tangente yo the graph at M_0 (x_0 ,y_0 ) is y=f^′ (x_0 )(x−x_0 )+f(x_0 ) ⇒f^′ (x_0 )x−f^′ (x_0 ).x_0 −y +f(x_0 )=0 ⇒ (8x_0 −4)x−(8x_0 −4)x_0 −y +f(x_0 )=0 the director vector is u^→ (1,8x_0 −4) and director vector of line 8x−y−6=0 is v^→ (1,8) we must have det(u^→ ,v^→ )=0 ⇒ determinant (((1 1)),((8x_0 −4 8)))=0 ⇒ 8−8x_0 +4 =0 ⇒2−2x_0 +1 =0 ⇒3−2x_0 =0 ⇒ x_0 =(3/2) ⇒f(x_0 )=4((3/2))^2 −4((3/2))+4 =9−6 +4 =7 ⇒ A((3/2),7)

$${we}\:{have}\:{f}\left({x}\right)=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}\:{the}\:{equation}\:{of}\:{tangente} \\ $$$${yo}\:{the}\:{graph}\:{at}\:{M}_{\mathrm{0}} \left({x}_{\mathrm{0}} ,{y}_{\mathrm{0}} \right)\:{is}\:{y}={f}^{'} \left({x}_{\mathrm{0}} \right)\left({x}−{x}_{\mathrm{0}} \right)+{f}\left({x}_{\mathrm{0}} \right) \\ $$$$\Rightarrow{f}^{'} \left({x}_{\mathrm{0}} \right){x}−{f}^{'} \left({x}_{\mathrm{0}} \right).{x}_{\mathrm{0}} −{y}\:+{f}\left({x}_{\mathrm{0}} \right)=\mathrm{0}\:\Rightarrow \\ $$$$\left(\mathrm{8}{x}_{\mathrm{0}} −\mathrm{4}\right){x}−\left(\mathrm{8}{x}_{\mathrm{0}} −\mathrm{4}\right){x}_{\mathrm{0}} −{y}\:+{f}\left({x}_{\mathrm{0}} \right)=\mathrm{0} \\ $$$${the}\:{director}\:{vector}\:{is}\:\overset{\rightarrow} {{u}}\left(\mathrm{1},\mathrm{8}{x}_{\mathrm{0}} −\mathrm{4}\right)\:{and}\:{director} \\ $$$${vector}\:{of}\:{line}\:\:\:\mathrm{8}{x}−{y}−\mathrm{6}=\mathrm{0}\:{is}\:\overset{\rightarrow} {{v}}\left(\mathrm{1},\mathrm{8}\right)\:{we}\:{must} \\ $$$${have}\:{det}\left(\overset{\rightarrow} {{u}},\overset{\rightarrow} {{v}}\right)=\mathrm{0}\:\Rightarrow\:\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{8}{x}_{\mathrm{0}} −\mathrm{4}\:\:\:\:\:\:\:\:\:\:\mathrm{8}}\end{vmatrix}=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{8}−\mathrm{8}{x}_{\mathrm{0}} +\mathrm{4}\:=\mathrm{0}\:\Rightarrow\mathrm{2}−\mathrm{2}{x}_{\mathrm{0}} +\mathrm{1}\:=\mathrm{0}\:\Rightarrow\mathrm{3}−\mathrm{2}{x}_{\mathrm{0}} =\mathrm{0}\:\Rightarrow \\ $$$${x}_{\mathrm{0}} =\frac{\mathrm{3}}{\mathrm{2}}\:\:\Rightarrow{f}\left({x}_{\mathrm{0}} \right)=\mathrm{4}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\mathrm{4} \\ $$$$=\mathrm{9}−\mathrm{6}\:+\mathrm{4}\:=\mathrm{7}\:\Rightarrow\:{A}\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{7}\right) \\ $$

Commented by liki last updated on 01/Dec/19

$$…{thanks}\:{sir}.. \\ $$

Commented by mathmax by abdo last updated on 01/Dec/19

$${you}\:{are}\:{welcome}. \\ $$

Answered by mr W last updated on 01/Dec/19

Q2: not clear what is zm_1 and m_4? please check the question. Q3: say one root is α, the other one is nα. α+nα=(n+1)α=−p α(nα)=nα^2 =q ⇒α=−(p/(n+1)) n×(p^2 /((n+1)^2 ))=q ⇒np^2 =(n+1)^2 q ⇒np^2 =n^2 q+2nq+q ⇒qn^2 +(2q−p^2 )n+q=0

$${Q}\mathrm{2}: \\ $$$${not}\:{clear}\:{what}\:{is}\:{zm\_}\mathrm{1}\:{and}\:{m\_}\mathrm{4}? \\ $$$${please}\:{check}\:{the}\:{question}. \\ $$$$ \\ $$$${Q}\mathrm{3}: \\ $$$${say}\:{one}\:{root}\:{is}\:\alpha,\:{the}\:{other}\:{one}\:{is}\:{n}\alpha. \\ $$$$\alpha+{n}\alpha=\left({n}+\mathrm{1}\right)\alpha=−{p} \\ $$$$\alpha\left({n}\alpha\right)={n}\alpha^{\mathrm{2}} ={q} \\ $$$$\Rightarrow\alpha=−\frac{{p}}{{n}+\mathrm{1}} \\ $$$${n}×\frac{{p}^{\mathrm{2}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }={q} \\ $$$$\Rightarrow{np}^{\mathrm{2}} =\left({n}+\mathrm{1}\right)^{\mathrm{2}} {q} \\ $$$$\Rightarrow{np}^{\mathrm{2}} ={n}^{\mathrm{2}} {q}+\mathrm{2}{nq}+{q} \\ $$$$\Rightarrow{qn}^{\mathrm{2}} +\left(\mathrm{2}{q}−{p}^{\mathrm{2}} \right){n}+{q}=\mathrm{0} \\ $$

Commented by liki last updated on 01/Dec/19

$$…{thank}\:{you}\:{very}\:{much}\:{sir}. \\ $$

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