已知二次函数y=x2﹣4x. (1) 在给出的直角坐标系内用描点法画出该二次函数的图象; (2) 根据所画的函数图象写出当x在什么范围内时,y≤0? (3) 根据所画的函数图象写出方程:x2﹣4x=5的解. 答案:解:y=x2﹣4x=(x﹣2)2﹣4,则抛物线的对称轴为直线x=﹣2,顶点坐标为(2,﹣4), 如图,
解:当0≤x≤4时,y≤0
解:由图象可知,x2﹣4x=5的解为x1=﹣1,x2=5—I'm thinking of the CCTV English Competition next week. I'm afraid I can't do very well. — . I'm sure you'll make it.
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