Question

find slope at p
find equation of tangent at point p

y = 6 - x^2
p (2,2)

Answer 1

im confused

Answer 2

confused about?

Answer 3

this homework

Answer 4

?

Answer 5

do you mean \[6-x^{2}\] to be the equation?

Answer 6

yes

Answer 7

First, find \(\frac{dy}{dx}\).\[\frac{dy}{dx}=0+2x^{2-1}\] or \[\frac{dy}{dx}=2x\] This is because the derivative of a constant is 0 and the derivitive of a power is \[\frac{d}{dx}x^{n}=nx^{n-1}\] Now just solve at \(x=2\):\[\frac{dy}{dx}=2(2)=4\] This is the slope, and you already know the tanget passes through \((2,2)\). Using point-slope form: \[y-y_{0}=m(x-x_{0})\] \[y-2=4(x-2)\] or, in terms of y: \[y=4x-6\]