Find the gradient of the tangent to the curve y=(3x-2)(x+4) at the points:a.(1,5) b.(0,-8) c.(-1,-15) 我是想求解答a.16 b.10 c.4可是我求的答案是 a.18 b.15 c.12 想知道正确答案是怎么得的这百度知道改版了?我输入了

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Findthegradientofthetangenttothecurvey=(3x-2)(x+4)atthepoints:a.(1,5)b.(0,-8)c.(-1,-15)我是想求解答a.16b.1

Find the gradient of the tangent to the curve y=(3x-2)(x+4) at the points:a.(1,5) b.(0,-8) c.(-1,-15) 我是想求解答a.16 b.10 c.4可是我求的答案是 a.18 b.15 c.12 想知道正确答案是怎么得的这百度知道改版了?我输入了
Find the gradient of the tangent to the curve y=(3x-2)(x+4) at the points:
a.(1,5) b.(0,-8) c.(-1,-15)
我是想求解答
a.16 b.10 c.4
可是我求的答案是 a.18 b.15 c.12 想知道正确答案是怎么得的
这百度知道改版了?我输入了标题按确定以为下一页会出现详细提问,结果直接就发上来了,我晕

Find the gradient of the tangent to the curve y=(3x-2)(x+4) at the points:a.(1,5) b.(0,-8) c.(-1,-15) 我是想求解答a.16 b.10 c.4可是我求的答案是 a.18 b.15 c.12 想知道正确答案是怎么得的这百度知道改版了?我输入了
楼主有学过导数么?将y=(3x-2)(x+4) 求导为 y=6x+10
分别将a ,b ,c 三点横坐标值代入导函数y=6x+10,即得过该点的切线斜率为
a.16 b.10 c.4

求y=(3x-2)(x+4)经过点a.(1,5) b.(0,-8) c.(-1,-15) 的切线斜率

解值域吗?

这就是求曲线在各点切线的斜率
原函数y=3x²+10x-8
求导,导函数y'=6x+10
因为他给的这3个点都在曲线上,直接把3个点的横坐标带入导函数中,得到的即为所求。

纠结中。。。。
题目。。。。。

这百度知道改版得让我也很晕,我跟一样想的,所以第一次也发错了,也就写在补充里

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