Find the possible values of a for which y=ax+1 is a tangent to y=3x^2-4x+4

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AS Core 1 For Edexcel - Pearson

Chapter 7: Graphs of function

Exercise 7C

Posted Sun 25th November, 2012 @ 12:14 by Mariya

1 Answer

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hello mariya,

to answer this question we first differentiate y=3x^2-4x-4 getting the gradient function 6x -4.

a= gradient function so y= (6x-4)(x) +1 

expand and we get y=6x^2-4x+1

next we need to find the coordinates of where the curve and the tangent meet so we do this.

6x^2 -4x +1=3x^2 -4x +1    which is = 3x^2 -3

expand to find x (3x-3)(x+1) so x is equal to 1 or -1

substitute them both into y= 3x^2 -4x +4 and we get y= 3 or 11

so our coordinates when both lines meet is (1,3) and (-1,11)

and if we substitute them into are tangent line (y=ax+1) then we can find the value for a 

3=a+1 and 11= -a+1 so a = 2 or -10

(I tend to make mistakes so be critical and point it out if i have)

either way hope this helps!

RTF

Answered Mon 26th November, 2012 @ 19:29 by Ragnaros the Firelord