The Student Room Group

c1 sketching curves

Question:
Find the possible values of a for whichy=ax+1 y = ax + 1 is a tangent to y=3x24x+4 y = 3x^{2} -4x + 4 .

Please look at my working
I still cant find a find to get to the answers of -10, 2

please help
Scan0030.jpg
Attachment not found


thank you in advance
Reply 1
Original post by bigmansouf
Question:
Find the possible values of a for whichy=ax+1 y = ax + 1 is a tangent to y=3x24x+4 y = 3x^{2} -4x + 4 .

Please look at my working
I still cant find a find to get to the answers of -10, 2

please help
Scan0030.jpg
Attachment not found


thank you in advance

If b=(4+a)b=-(4+a)

then b2=[(4+a)]2=(4+a)2b^2 = \left[-(4+a)\right]^2 = (4+a)^2

Just like e.g. (3)2=32(-3)^2 = 3^2
Original post by bigmansouf
Question:
Find the possible values of a for whichy=ax+1 y = ax + 1 is a tangent to y=3x24x+4 y = 3x^{2} -4x + 4 .

Please look at my working
I still cant find a find to get to the answers of -10, 2


When you were finding b^2, you made a mistake.
I think you should expand the brackets first, - (4+a) first = -4 + 4a and then square them.
Or, do exactly what notnek said - think of it like that, if you find it easier - rearrange the 4 + a to a + 4 so it matches (x+a)^2 and solve it like that.
Reply 3
Original post by Chittesh14
When you were finding b^2, you made a mistake.
I think you should expand the brackets first, - (4+a) first = -4 + 4a and then square them.
Or, do exactly what notnek said - think of it like that, if you find it easier - rearrange the 4 + a to a + 4 so it matches (x+a)^2 and solve it like that.


Original post by notnek
If b=(4+a)b=-(4+a)

then b2=[(4+a)]2=(4+a)2b^2 = \left[-(4+a)\right]^2 = (4+a)^2

Just like e.g. (3)2=32(-3)^2 = 3^2

thank you i looked through my work and realised i did a mistake
Original post by bigmansouf
thank you i looked through my work and realised i did a mistake


No problem :smile:.

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