What is the tangent of f(x)=(x-3)^2-2,\at x=0 ?

The tangent of f(x)=(x-3)^2-2,\at x=0 is y=-6x+7

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tangent of f(x)=(x-3)^2-2,\at x=0

tangent of

f (x) = (x - 3)^{2} - 2, at x = 0

y = - 6 x + 7

Solution steps

Find the tangent point:

(0, 7)

Find the slope of

f (x) = (x - 3)^{2} - 2 : \frac{df ( x )}{d x} = 2 x - 6

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 6

Find the line with slope m=

- 6

and passing through

(0, 7) : y = - 6 x + 7

y = - 6 x + 7

\frac{1}{x} y^{^{'}} = \frac{y}{x ^{2}} - \frac{1}{y}

\frac{\partial}{\partial y} (y^{3} + x^{2} y - 3 y)

n = 1 \sum \infty \frac{1}{n ln ( n )}

\int \frac{cos ( θ )}{9 - sin ^{2} ( θ )} d θ

\frac{d}{d x} (\frac{y}{x ^{2} + y ^{2}})

What is the tangent of f(x)=(x-3)^2-2,\at x=0 ?
The tangent of f(x)=(x-3)^2-2,\at x=0 is y=-6x+7