What is the tangent of f(x)=x^2-5,(-3,4) ?

The tangent of f(x)=x^2-5,(-3,4) is y=-6x-14

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tangent of f(x)=x^2-5,(-3,4)

tangent of

f (x) = x^{2} - 5, (- 3, 4)

y = - 6 x - 14

Solution steps

Find the slope of

f (x) = x^{2} - 5 : \frac{df ( x )}{d x} = 2 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 = - 6

Find the line with slope m=

- 6

and passing through

(- 3, 4) : y = - 6 x - 14

y = - 6 x - 14

\int (x - 3)^{2} \cdot e^{- x} d x

\int \frac{1}{( x - 2 ) ( x + 2 ) ^{2}} d x

\int \frac{1}{x ^{2}} 1 - \frac{1}{x} d x

\frac{d y}{d x} = y cos (3 x + 2)

\frac{d y}{d t} + 7 y = e^{5 t}

What is the tangent of f(x)=x^2-5,(-3,4) ?
The tangent of f(x)=x^2-5,(-3,4) is y=-6x-14