maths questions-transformation on the graphs of function(3)

2008-10-16 5:51 am

22.The figure shows the graph of y=f(x), where A and B are x-intercepts and C is the vertex.圖內:有一條parabola，open downwards既，有A，B2點，係y=f(x)既解，y-intercept 同另一個parabloay=g(x)既一樣，形狀一樣，x-intercept係D，E2點。
Consider f(x)= -x^2-8x-12.
(a)(i) Find the y-intercept of the graph of y=f(x). Ans:-12
(ii)Express y=f(x) in the form of y=a(x-h)^2 +k Ans:-(x+4)^2 +4
(iii) Hence find the coordinates of the vertex of y=f(x) Ans: (-4,4)
上面果D我知點計，但我下面果D唔知點計，根據上題答案，解答下列題目，2題有關架。
Now y=f(x) is translates k units to the right to become y=g(x), such that they have the same y-intercept.
(a)(i) Find the value of k.' Ans:8
(ii) Express y=g(x) in the form y=ax^2 +bx+c.
y=g(x)y=g(x)
Ans: y= -x^2+8x-12
唔識列式教教我，多謝!

回答 (1)

用戶10012

2008-10-16 6:01 am

✔ 最佳答案

y = f(x) = -(x+4)^2 + 4.
When translate to the right by k units, y becomes g(x), that is
y = g(x) = - ( x + 4 - k)^2 + 4................(1)
When x = 0, f(x) = - 16 + 4 = - 12. That is the y intercept of g(x) is also -12.
Put x = 0, g(x) = -(4-k)^2 + 4 = -12
-k^2 - 16 + 8k + 4 + 12 = 0
k^2 - 8k = 0
k(k - 8) = 0
k = 0 (rej. because translation is not zero) and k = 8.
Sub. into (1), we get
y = g(x) = -(x + 4 - 8)^2 + 4
= -(x-4)^2 + 4
= -x^2 + 8x - 12.

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