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# Maths Question watch

1. The line y=4x+k is a tangent to the curve y=x^2+4x+3 at the point P

Find the value of k
2. Solve as a simultaneous equation, y = y method, as you're looking for them to meet.

This will then involve factorising - can you identify P from there? How do you know it's a tangent at that point, rather than just a point of intersection?
3. Scratch that. The second step only works easily when k is already known. Do the first step, then ask yourself what condition needs to be met for this to be a tangent?

Spoiler:
Show
b^2-4ac=0 must be satisfied
4. why dont you just differentiate the quadratic equation and equate it to 4. Then solve for x. Sub this value into the original quadratic and get the y value. then sub both the x and y values in the linear equation to get k.
5. Im really confuse with this question, could someone show me some working out?
6. Equate them to get x^2+3-k=0 then since there is one point, P, the discriminant=0. So b^2-4ac=0 12-4k=0 therefore k=3 and so x=0 thus y=3 k=3 P=(0,3)
7. (Original post by haken102)
Im really confuse with this question, could someone show me some working out?
Okay.
Spoiler:
Show

y=4x+k
y=x^2+4x+3

So equate them...

x^2+4x+3 = 4x+k
x^2 + (3-k) = 0

So for b^2-4ac, a=1, b=0, c=(3-k)

b^2-4ac=0 gives a tangent

(0)^2-4(1)(3-k)=0
-4(3-k)=0
3-k=0
3=k

8. why did you use the discriminant?
9. nvm, i figured it out. Thank you so much guys.
10. (Original post by haken102)
nvm, i figured it out. Thank you so much guys.
Next time you will be better off posting in the maths forum

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