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Help me to do the function problems in the first volume of Grade Two Mathematics! ! ! ! ! ! !
1, ① let this resolution function be: y=kx+b,

Substitute x=-3, y=2, x= 1 and y=6 into the above formulas respectively, and obtain

2=-3k+b - ①

6=k+b - ②

②-① get 4k=4.

k= 1

Substitute k= 1 into ② to get b=5.

So the resolution function is: y=x+5.

② Let x=0, and you can get y = 5;; Let y=0 and x=-5.

Therefore, the required triangle area is

1/2×5×∣-5∣= 12.5

2.① Because the image of function y=(2m= 1)x+m-3 passes through the origin,

So, m-3=0, m=3.

② Because the image of function y=(2m= 1)x+m-3 does not pass through the second quadrant,

Therefore, 2m- 1 > 0, m-3≤0 ∴ 1/2≤m≤3.

Note: 1/2 is half.