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can you explain to me how the x intercept of each problem become like that. The one that encircle in each problem is part that I don't understand, can you give me the full step and explanation

can you explain to me how the x intercept of each problem become like that. The one that encircle in each problem is part that I don't understand, can you give me the full step and explanation

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Hi can you explain to me how the x intercept of each problem become like that. The one that encircle in each problem is part that I don't understand, can you give me the full step and explanation thank you!
2) At what peaist do the ペー! curve ye >(xキ-) hawe a ベヤ slopes of 2? スー Curve: 2(x*+ ax+1) ス*+2×t = 1 =2 Xこ0 Xヤスニ0 1- キメ dy (ス)素(スー)) - タム) 沖 (xti) ズニー2 ニ **+ 2x = 0 ス ー(スー) ×(メ+2)=0 ニ iス=o iズ=-2 X-1 2. X-1 ス+1-メt (メン 0-1 -ス-1 dy こ ; M=ス= ース ニ (スキア -3 2 Cス) y = -1 y = 3 2(スt- 2 Pz (~2,3)
Transcribed Image Text:2) At what peaist do the ペー! curve ye >(xキ-) hawe a ベヤ slopes of 2? スー Curve: 2(x*+ ax+1) ス*+2×t = 1 =2 Xこ0 Xヤスニ0 1- キメ dy (ス)素(スー)) - タム) 沖 (xti) ズニー2 ニ **+ 2x = 0 ス ー(スー) ×(メ+2)=0 ニ iス=o iズ=-2 X-1 2. X-1 ス+1-メt (メン 0-1 -ス-1 dy こ ; M=ス= ース ニ (スキア -3 2 Cス) y = -1 y = 3 2(スt- 2 Pz (~2,3)
4) Find the eqeation of the tamgent Line and normal lime to the uerue whose norml ine is jaaraled to the Cine ex+3y-4=o, y = Ja-3 shetch the graph. Curve: yeVメー3 then NL /L:G×+3y- 4 =0 ; mL=ー 3 MNL こ Mr=-2 ; Mnこー Mar こ- dy こ Equ. gthe TL mouoe P(4.り e P(x,.y.) f ス、=4 '3 y-y, zm (x-x,) - JA -3 y-1į (x-4) ー2 2 ニ Egá of the wC Masc= -2 @ P(4.1) J-yr = mCx-π.) 2 x-3 2メ-3 Y, = J-1z(ォ-4) 三 ニ 2 y-1= -2(x-4) y, = | PC 4,) 24 -2= ズー .4. %3D Jス-3 ズースy-ス= 2x+y-9=0 メー3 =} X,= 4 NL
Transcribed Image Text:4) Find the eqeation of the tamgent Line and normal lime to the uerue whose norml ine is jaaraled to the Cine ex+3y-4=o, y = Ja-3 shetch the graph. Curve: yeVメー3 then NL /L:G×+3y- 4 =0 ; mL=ー 3 MNL こ Mr=-2 ; Mnこー Mar こ- dy こ Equ. gthe TL mouoe P(4.り e P(x,.y.) f ス、=4 '3 y-y, zm (x-x,) - JA -3 y-1į (x-4) ー2 2 ニ Egá of the wC Masc= -2 @ P(4.1) J-yr = mCx-π.) 2 x-3 2メ-3 Y, = J-1z(ォ-4) 三 ニ 2 y-1= -2(x-4) y, = | PC 4,) 24 -2= ズー .4. %3D Jス-3 ズースy-ス= 2x+y-9=0 メー3 =} X,= 4 NL
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