# c1 differentiation tangentsWatch

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#1
hey guys this question is given me a headache can some one help me please?

find the coordinates of the following curves where the gradient has the stated value.

y=x^3 - 6x^2 + 10x + 5 gradient -2

so this how i started it

3x^2 - 12x + 10 = -2
3x^2 - 12 x = -12

and then this is where i get really confused can some help me please?
0
5 years ago
#2
(Original post by mrdetermination)
hey guys this question is given me a headache can some one help me please?

find the coordinates of the following curves where the gradient has the stated value.

y=x^3 - 6x^2 + 10x + 5 gradient -2

so this how i started it

3x^2 - 12x + 10 = -2
3x^2 - 12 x = -12

and then this is where i get really confused can some help me please?
.

What kind of equation is that?
1
#3
i though it was a quadratic equation then you change is in y=mx+c then solve if im using the wrong method can you show me the right one
0
5 years ago
#4
(Original post by mrdetermination)
i though it was a quadratic equation then you change is in y=mx+c then solve if im using the wrong method can you show me the right one
It is a quadratic equation. Just solve it.
0
5 years ago
#5
all you need to do is solve that quadratic equation to find x, then plug those values into the original equation to get the y coordinates
0
#6
okay thx guys
0
5 years ago
#7
(Original post by mrdetermination)
hey guys this question is given me a headache can some one help me please?

find the coordinates of the following curves where the gradient has the stated value.

y=x^3 - 6x^2 + 10x + 5 gradient -2

so this how i started it

3x^2 - 12x + 10 = -2
3x^2 - 12 x = -12

and then this is where i get really confused can some help me please?
It is a quadratic equation and it can be factorised. SO factorise it to find your x co ordinates, and plug those back into the orignal curve function to find the corresponding y co ordinates
0
5 years ago
#8
.divide everything by 3 to get x^2-4x +4.

Thats obviously a double root at x-2.
0
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