中四數學題

1/
h=50+16t-4t²
h=-4t²+16t+50
h=-4(t²-4)+50
h=-4(t-2)²+66
2秒後該物體達到最高處



g(x)=x²+2

h(x)=(x-3)²+2
......=x²-6x+9+2
......=x²-6x+11

某人於一高50米的大廈天台斜拋一物体,於7秒後該物体離地面的垂直高度為h米,而h和t的函數關係式為h=50+16t-4t^2。
經過多少秒該物體達到最高處?
Since h = 50+16t-4t^2
h = -4t^2+16t+50
h = -4t^2+16t-16+66
h = -4(t^2-4t+4)+66
h = -4(t-2)^2+66
when t = 2 , h attains its maximum value 66
So , The object attains the maximum height 66m after throwing 2s.

把函數y=x^2的圖像向上平移2單位後得函數y=g(x)的圖像。隨後又把函數y=g(x)的圖像向右平移3單位,得另一函數y=h(x)的圖像。試寫岀函數g(x)和h(x),並化簡h(x)。
Because y = g(x) is obtained by translating y = x^2 in the positive direction of y-axis by 2 units and y = h(x) is obtained by translating y = g(x) in the positive direction of x-axis by 3 units ,
g(x) = x^2+2
h(x) = g(x-3)
= (x-3)^2+2
= x^2-6x+11