is this correct?

[1141]

Hey there

I was doing this question:

The normal to the curve with equation y-x[sup:2n46xnx8]2[/sup:2n46xnx8] at the point for which x=2 meets the curve again at p. Fin dhte co-ordinates of p.

This is what I've done so for:

y-x[sup:2n46xnx8]2[/sup:2n46xnx8]
f'(x) = -2x

y-x[sup:2n46xnx8]2[/sup:2n46xnx8]=0
y-(2[sup:2n46xnx8]2[/sup:2n46xnx8])=0
y-4=0
y=4

y-4=-2x(x-2)
y-4=-2x[sup:2n46xnx8]2[/sup:2n46xnx8]+4x
y=-2x[sup:2n46xnx8]2[/sup:2n46xnx8]+4x+4

then I substituted y into y-x[sup:2n46xnx8]2[/sup:2n46xnx8]

y-x[sup:2n46xnx8]2[/sup:2n46xnx8]
-2x[sup:2n46xnx8]2[/sup:2n46xnx8]+4x+4-x[sup:2n46xnx8]2[/sup:2n46xnx8]
-3x[sup:2n46xnx8]2[/sup:2n46xnx8]+4x+4

Then I used the formula to find the roots of that quadratic coming to the answer that,
x=3
or
x= -1

That's where I stopped. I looked at the question again and became confused. "The normal to the curve with equation y-x[sup:2n46xnx8]2[/sup:2n46xnx8]" confused me. Is y-x[sup:2n46xnx8]2[/sup:2n46xnx8] the equation for the normal or for the curve?? And the working out that I did, am I on the right track?


Thanks.

 

[BigGlenntheHeavy]

\(\displaystyle y-x^{2} \ is \ not \ an \ equation, \ however \ I \ will \ assume \ that \ f(x) \ = \ -x^{2}.\)

\(\displaystyle Then, \ here \ is \ the \ graph, \ you \ should \ be \ able \ to \ take \ it \ from \ here.\)

[attachment=1:3l195m95]def.jpg[/attachment:3l195m95]

\(\displaystyle However, \ if \ f(x) \ = \ x^{2}, \ then \ this \ is \ the \ graph.\)

[attachment=0:3l195m95]ghi.jpg[/attachment:3l195m95]