What is the tangent of f(x)=-4x^2+6,\at x=-1 ?

The tangent of f(x)=-4x^2+6,\at x=-1 is y=8x+10

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tangent of f(x)=-4x^2+6,\at x=-1

tangent of

f (x) = - 4 x^{2} + 6, at x = - 1

y = 8 x + 10

Solution steps

Find the tangent point:

(- 1, 2)

Find the slope of

f (x) = - 4 x^{2} + 6 : \frac{df ( x )}{d x} = - 8 x

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 = 8

Find the line with slope m=

8

and passing through

(- 1, 2) : y = 8 x + 10

y = 8 x + 10

\frac{d y}{d x} = - \frac{( x e ^{- x^{2}} )}{y}

\int x^{(\frac{3}{2})} d x

\frac{d}{d x} (arcsin (x) \frac{8}{π})

n = 1 \sum \infty n \cdot \frac{1}{2 ^{n}}

\int \frac{1}{x ^{2} + 8 x} d x

What is the tangent of f(x)=-4x^2+6,\at x=-1 ?
The tangent of f(x)=-4x^2+6,\at x=-1 is y=8x+10