Let-p-and-Q-be-points-on-the-curve-y-x-2-2x-while-x-2-and-x-2-h-respectively-Epress-the-gradient-of-PQ-in-terms-of-h-

Tinku Tara June 4, 2023 Relation and Functions 0 Comments

Question Number 129052 by oustmuchiya@gmail.com last updated on 12/Jan/21

Let p and Q be points on the curve y=x^2 −2x while x=2 and x=2+h respectively. Epress the gradient of PQ in terms of h.

$${Let}\:\boldsymbol{\mathrm{p}}\:{and}\:\boldsymbol{\mathrm{Q}}\:{be}\:{points}\:{on}\:{the}\:{curve} \\ $$$$\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}\:{while}\:\boldsymbol{{x}}=\mathrm{2}\:{and}\:\boldsymbol{{x}}=\mathrm{2}+\boldsymbol{{h}} \\ $$$$\boldsymbol{{respectively}}.\:\boldsymbol{{E}}{press}\:{the}\:{gradient} \\ $$$${of}\:\boldsymbol{\mathrm{P}}{Q}\:{in}\:{terms}\:{of}\:\boldsymbol{{h}}. \\ $$

Commented by benjo_mathlover last updated on 12/Jan/21

P(2,0); Q(2+h,h^2 +2h) gradient of PQ = ((h^2 +2h−0)/(2+h−2))=((h^2 +2h)/h)=h+2

$$\mathrm{P}\left(\mathrm{2},\mathrm{0}\right);\:\mathrm{Q}\left(\mathrm{2}+\mathrm{h},\mathrm{h}^{\mathrm{2}} +\mathrm{2h}\right) \\ $$$$\mathrm{gradient}\:\mathrm{of}\:\mathrm{PQ}\:=\:\frac{\mathrm{h}^{\mathrm{2}} +\mathrm{2h}−\mathrm{0}}{\mathrm{2}+\mathrm{h}−\mathrm{2}}=\frac{\mathrm{h}^{\mathrm{2}} +\mathrm{2h}}{\mathrm{h}}=\mathrm{h}+\mathrm{2} \\ $$

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