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Maths Qusetion!Urgent
發問:
25 The figure shows the graph of a quadratic equation y= f(x) with P (p,q) as itsvertex.The graph cuts the x-axis at A(1,0) and B(5,0), cuts the y-axis at C(0,10)and passes through R(6,r).a)Find f(x) ans: f(x) = 2x^2 -12x +10b) Find the values of p, q and r ans: p = 3, q= -8 , r =10c)It... 顯示更多 25 The figure shows the graph of a quadratic equation y= f(x) with P (p,q) as its vertex.The graph cuts the x-axis at A(1,0) and B(5,0), cuts the y-axis at C(0,10) and passes through R(6,r). a)Find f(x) ans: f(x) = 2x^2 -12x +10 b) Find the values of p, q and r ans: p = 3, q= -8 , r =10 c)It is given that the graph of another quadratic function y=g(x) passes through P and has R as its vertex i) Express g(x) in the form of a(x-h)^2 + k ans:-2(x-6)^2 +10 ******(this part i dunno how to do)ii) If Q is a point on the graph of y= g(x) such that QR=PR,show that QR is parallel to PC The figure is here: http://s457.photobucket.com/user/aldikwok1996/media/CAM00040.jpg.html?sort=3&o=0 Just answer one part c) ii) only
最佳解答:
Ting Yan, 你好。 首先，用番你之前得出的結果。 P = (3, -8) R = (6, 10) g(x) = -2(x-6)2 +10 Let Q = (a,b). Then, you know that g(a) = b, because Q is a point on y = g(x). That is, b = -2(a-6)2 +10 ...[1] On the other hand, QR = PR QR2 = PR2 (a-6)2+(b-10)2 = (3-6)2+(-8-10)2 = 333 From [1], we know that (a-6)2 = (b-10)/(-2) Therefore, (b-10)/(-2) + (b-10)2 = 333 (b-10) - 2(b-10)2 = -666 b-10 - 2(b2-20b+100) = -666 b-10 - 2b2 +40b -200 = -666 -2b2 +41b -210 = -666 2b2 -41b -456 = 0 (b+8)(2b-57)=0 b = -8 or 57/2 (rejected, since max of b is 10, by [1].) (a-6)2 = (-8-10)/(-2) = 9 a-6 = ±3 a = 6±3 = 3 or 9 Note that P = (3, -8), so Q = (9, -8). Slope of QR = (10+8)/(6-9) = -6 Slope of PC = (10+8)/(0-3) = -6 Therefore, QR//PC. 計就係咁計，因為要寫清楚ｄ steps。 但其實你諗的話，一早都可以知。 你望下幅圖。 三角形PRQ係一個等腰三角形（因為對稱、R是vertex）， 所以你一早就知道Q的 y-coordinate係q = -10。 而且，整個 CPQR 根本就是一個平行四邊形。
其他解答:
Maths Qusetion!Urgent
發問:
25 The figure shows the graph of a quadratic equation y= f(x) with P (p,q) as itsvertex.The graph cuts the x-axis at A(1,0) and B(5,0), cuts the y-axis at C(0,10)and passes through R(6,r).a)Find f(x) ans: f(x) = 2x^2 -12x +10b) Find the values of p, q and r ans: p = 3, q= -8 , r =10c)It... 顯示更多 25 The figure shows the graph of a quadratic equation y= f(x) with P (p,q) as its vertex.The graph cuts the x-axis at A(1,0) and B(5,0), cuts the y-axis at C(0,10) and passes through R(6,r). a)Find f(x) ans: f(x) = 2x^2 -12x +10 b) Find the values of p, q and r ans: p = 3, q= -8 , r =10 c)It is given that the graph of another quadratic function y=g(x) passes through P and has R as its vertex i) Express g(x) in the form of a(x-h)^2 + k ans:-2(x-6)^2 +10 ******(this part i dunno how to do)ii) If Q is a point on the graph of y= g(x) such that QR=PR,show that QR is parallel to PC The figure is here: http://s457.photobucket.com/user/aldikwok1996/media/CAM00040.jpg.html?sort=3&o=0 Just answer one part c) ii) only
最佳解答:
Ting Yan, 你好。 首先，用番你之前得出的結果。 P = (3, -8) R = (6, 10) g(x) = -2(x-6)2 +10 Let Q = (a,b). Then, you know that g(a) = b, because Q is a point on y = g(x). That is, b = -2(a-6)2 +10 ...[1] On the other hand, QR = PR QR2 = PR2 (a-6)2+(b-10)2 = (3-6)2+(-8-10)2 = 333 From [1], we know that (a-6)2 = (b-10)/(-2) Therefore, (b-10)/(-2) + (b-10)2 = 333 (b-10) - 2(b-10)2 = -666 b-10 - 2(b2-20b+100) = -666 b-10 - 2b2 +40b -200 = -666 -2b2 +41b -210 = -666 2b2 -41b -456 = 0 (b+8)(2b-57)=0 b = -8 or 57/2 (rejected, since max of b is 10, by [1].) (a-6)2 = (-8-10)/(-2) = 9 a-6 = ±3 a = 6±3 = 3 or 9 Note that P = (3, -8), so Q = (9, -8). Slope of QR = (10+8)/(6-9) = -6 Slope of PC = (10+8)/(0-3) = -6 Therefore, QR//PC. 計就係咁計，因為要寫清楚ｄ steps。 但其實你諗的話，一早都可以知。 你望下幅圖。 三角形PRQ係一個等腰三角形（因為對稱、R是vertex）， 所以你一早就知道Q的 y-coordinate係q = -10。 而且，整個 CPQR 根本就是一個平行四邊形。
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