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關於中二MATHS題!(十五分)
發問:
Show your steps!題1: Find the coordiantes of the point of intersection of the parabola y=x^2 and the line y=2x-1題2: Find the x-intercepts of the parabola y=x^2-5x+3題3: Find the x-intercept of the parabola y=x^2-2(開方2)x+2題4: Find the coordinates of the points of intersection of the... 顯示更多 Show your steps! 題1: Find the coordiantes of the point of intersection of the parabola y=x^2 and the line y=2x-1 題2: Find the x-intercepts of the parabola y=x^2-5x+3 題3: Find the x-intercept of the parabola y=x^2-2(開方2)x+2 題4: Find the coordinates of the points of intersection of the parabola y=x^2 and y=x+3 題5: Find the corodinates of the points of intersection of the parabola y=x^2-ax and the line y=bx-ab
題1: Find the coordiantes of the point of intersection of the parabola y=x2 and the line y=2x-1 y=x^2___________(1) y=2x-1__________(2) put(2)into(1), 2x-1=x2 x2-2x+1=0 (x-1)^2=0 x=1 put x=1 into (2) y=2(1)-1 =1 the coordinates=(1,1) 題2: Find the x-intercepts of the parabola y=x2-5x+3 put y=0 into y=x2-5x+3 0=x2-5x+3 x2-5x+25/4=13/4 (x-5/2)^2=13/4 x-5/2=(√13)/2 x=(5√13)/2 x-intercepts of the parabola =(5√13)/2 題3: Find the x-intercept of the parabola y=x2-2(√2)x+2 put y=0 into y=x2-2(√2)x+2 0=x2-2√2x+2 (x-√2)^2=0 x=√2 x-intercepts of the parabola =√2 題4: Find the coordinates of the points of intersection of the parabola y=x2 and y=x+3 y=x2__________________(1) y=x+3_________________(2) put (2) into (1), x2=x+3 x2-x=3 x2-x+1/4=13/4 (x-1/2)^2=13)/4 x-1/2=(√13)/2 x=(1√13)/2 put x=(1√13)/2 into y=x+3 y=(7√13)/2 the coordinates=[(1√13)/2 ,(7√13)/2] 題5: Find the corodinates of the points of intersection of the parabola y=x2-ax and the line y=bx-ab y=x2_____________(1) y=bx-ab __________(2) put (2) into(1), x2-ax=bx-ab x2-(a+b)x+ab=0 (x-a)(x-b)=0 x=a or x=b put x=a into(2), y=ab-ab =0 put x=b into(2), y=b2-ab the corodinates =(a,0) or(b,b2-ab)
其他解答:
(1) y= 2x - 1 -----(1) y = x^2 ------(2) sub Eq(1) into Eq(2): x^2 - 2x + 1 = 0 (x -1)^2 =0 when x = 1, y = 2(1) - 1 = 1 The intersection coordinates are (1,1) (2) y = x^2 - 5x + 3 the curve cuts x - axis which is the x-intercept set y = 0, x^2 - 5x + 3 = 0 x = {5 + √[5^2 - 4(1)(3)]} /2 = 2.5 + 0.5√13 x = {5 - √[5^2 - 4(1)(3)]} /2 = 2.5 - 0.5√13 The x intercepts are (2.5 + 0.5√13, 0) and (2.5 - 0.5√13, 0) (3) the curve cuts x-axis which are the x intercepts set y = 0, x^2 - 2√2x + 2 = 0 (x - √2)^2 = 0 x = √2 The x - intercept is ( √2, 0) (4) y = x + 3 -------(1) y = x^2 -------(2) sub Eq(1) into Eq(2): x^2 = x + 3 x^2 - x - 3 = 0 x = {1 + √[(1)^2 - 4(1)(-3)]}/2 = 0.5 + 0.5√13 x = {1 - √[(1)^2 - 4(1)(-3)]}/2 = 0.5 - 0.5√13 The intersections are (0.5 + 0.5√13, 0) and (0.5 - 0.5√13, 0) (5) y = x^2 - ax ------(1) y = bx - ab ------(2) sub Eq(1) into Eq(2): x^2 - ax = bx - ab x^2 - (a+b)x + ab = 0 (x - a)(x - b) = 0 x = a, or x= b when x = a, y = a^2 - a^2 = 0 when x = b, y = b^2 - ab The intersection coordinates are (0, a) and (b, b^2 - ab)
關於中二MATHS題!(十五分)
發問:
Show your steps!題1: Find the coordiantes of the point of intersection of the parabola y=x^2 and the line y=2x-1題2: Find the x-intercepts of the parabola y=x^2-5x+3題3: Find the x-intercept of the parabola y=x^2-2(開方2)x+2題4: Find the coordinates of the points of intersection of the... 顯示更多 Show your steps! 題1: Find the coordiantes of the point of intersection of the parabola y=x^2 and the line y=2x-1 題2: Find the x-intercepts of the parabola y=x^2-5x+3 題3: Find the x-intercept of the parabola y=x^2-2(開方2)x+2 題4: Find the coordinates of the points of intersection of the parabola y=x^2 and y=x+3 題5: Find the corodinates of the points of intersection of the parabola y=x^2-ax and the line y=bx-ab
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最佳解答:題1: Find the coordiantes of the point of intersection of the parabola y=x2 and the line y=2x-1 y=x^2___________(1) y=2x-1__________(2) put(2)into(1), 2x-1=x2 x2-2x+1=0 (x-1)^2=0 x=1 put x=1 into (2) y=2(1)-1 =1 the coordinates=(1,1) 題2: Find the x-intercepts of the parabola y=x2-5x+3 put y=0 into y=x2-5x+3 0=x2-5x+3 x2-5x+25/4=13/4 (x-5/2)^2=13/4 x-5/2=(√13)/2 x=(5√13)/2 x-intercepts of the parabola =(5√13)/2 題3: Find the x-intercept of the parabola y=x2-2(√2)x+2 put y=0 into y=x2-2(√2)x+2 0=x2-2√2x+2 (x-√2)^2=0 x=√2 x-intercepts of the parabola =√2 題4: Find the coordinates of the points of intersection of the parabola y=x2 and y=x+3 y=x2__________________(1) y=x+3_________________(2) put (2) into (1), x2=x+3 x2-x=3 x2-x+1/4=13/4 (x-1/2)^2=13)/4 x-1/2=(√13)/2 x=(1√13)/2 put x=(1√13)/2 into y=x+3 y=(7√13)/2 the coordinates=[(1√13)/2 ,(7√13)/2] 題5: Find the corodinates of the points of intersection of the parabola y=x2-ax and the line y=bx-ab y=x2_____________(1) y=bx-ab __________(2) put (2) into(1), x2-ax=bx-ab x2-(a+b)x+ab=0 (x-a)(x-b)=0 x=a or x=b put x=a into(2), y=ab-ab =0 put x=b into(2), y=b2-ab the corodinates =(a,0) or(b,b2-ab)
其他解答:
(1) y= 2x - 1 -----(1) y = x^2 ------(2) sub Eq(1) into Eq(2): x^2 - 2x + 1 = 0 (x -1)^2 =0 when x = 1, y = 2(1) - 1 = 1 The intersection coordinates are (1,1) (2) y = x^2 - 5x + 3 the curve cuts x - axis which is the x-intercept set y = 0, x^2 - 5x + 3 = 0 x = {5 + √[5^2 - 4(1)(3)]} /2 = 2.5 + 0.5√13 x = {5 - √[5^2 - 4(1)(3)]} /2 = 2.5 - 0.5√13 The x intercepts are (2.5 + 0.5√13, 0) and (2.5 - 0.5√13, 0) (3) the curve cuts x-axis which are the x intercepts set y = 0, x^2 - 2√2x + 2 = 0 (x - √2)^2 = 0 x = √2 The x - intercept is ( √2, 0) (4) y = x + 3 -------(1) y = x^2 -------(2) sub Eq(1) into Eq(2): x^2 = x + 3 x^2 - x - 3 = 0 x = {1 + √[(1)^2 - 4(1)(-3)]}/2 = 0.5 + 0.5√13 x = {1 - √[(1)^2 - 4(1)(-3)]}/2 = 0.5 - 0.5√13 The intersections are (0.5 + 0.5√13, 0) and (0.5 - 0.5√13, 0) (5) y = x^2 - ax ------(1) y = bx - ab ------(2) sub Eq(1) into Eq(2): x^2 - ax = bx - ab x^2 - (a+b)x + ab = 0 (x - a)(x - b) = 0 x = a, or x= b when x = a, y = a^2 - a^2 = 0 when x = b, y = b^2 - ab The intersection coordinates are (0, a) and (b, b^2 - ab)
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