1. ## finding quadratics with working

This is Q25 form Cambridge Mathematics year 11 3 unit (in case anyone has the book rather than following me)

(a) show that the tangent to P:y=ax^2+bx+c with gradient m has y-intercept c-[(m-b)^2]/4a.

y'=2ax+b, which is m

rearrange for x=m-b/2a

sub x into y gives y=a(m-b/2a)^a+b(m-b/2a)+c

then I used the gradient/point formula and the above values for m, x and y which simplifed to give y-intercept c-[(m-b)^2]/4a.

So part a is done

(b) Hence find any equations of any quadratics that pass through the origin and are tangents to both y=-2x-4 and y=8x-49.

y=-2x-4, y'=-2x therefore m=-2 and c=0 (1)
y=8x-49, y'=8 therefore m=8 and c=0 (2)

I sub'ed the values from (1) and (2) into y-intercept=c-[(m-b)^2]/4a and got
16a=b^2+4b+4 (1)
196a=b^2-16b-64 (2)

and then solved simultaneous equations and found b=0 or 6
Now sub b=6 back onto (1) or (2) gives a=1 which gives

y=x^2-6x and that is one of the answers

Thank you.

2. Originally Posted by slaypullingcat
(b) Hence find any equations of any quadratics that pass through the origin and are tangents to both y=-2x-4 and y=8x-49.

y=-2x-4, y'=-2x therefore m=-2 and c=0 (1)
y=8x-49, y'=8 therefore m=8 and c=0 (2)

I sub'ed the values from (1) and (2) into y-intercept=c-[(m-b)^2]/4a and got
16a=b^2+4b+4 (1)
196a=b^2-16b+64 (2)
Everything is correct up to there, except that the constant term in that last equation should be +64, not –64. Now multiply (1) by 49, (2) by 4, and subtract. You get $0 = 49(b^2+4b+4) - 4(b^2-16b+64)$, which simplifies to $45b^2 + 260b - 60 = 0$. Divide through by 5, and then factorise it as $(b+6)(9b-2) = 0$, with solutions b = -6 and 2/9, as required.

3. Thank you so much!