Find value of p for y= x^2 + px + 2 so gradient m = -4 at x = 3

[Sirus]

Hi can anyone help me to do this question?:


The equation for a parabola is given by y= x^2+px+2 determine the value of p so that the gradient of the parabola is -4 at the point where x = 3.

 

[Sirus]

Never mind i found it out myself.

Let m = the gradient of the tangent


y=x^2+px+2
m=-4
y'=2x+p
-4=2(3)+p
-4=6+p
-4-6=p
p=-10

y=x^2-10x+2
y'=2x-10
y'(3)=m=2(3)-10
y'(3)=m = 6-10
m=-4

 

[stapel]

Never mind i found it out myself.

Let m = the gradient of the tangent


y=x^2+px+2
m=-4
y'=2x+p
-4=2(3)+p
-4=6+p
-4-6=p
p=-10

y=x^2-10x+2
y'=2x-10
y'(3)=m=2(3)-10
y'(3)=m = 6-10
m=-4

Thank you (1) for coming back and posting your work for other students to see, and (2) for typing it out so nicely!