What is the tangent of f(x)=x^2+3,(-5,28) ?

The tangent of f(x)=x^2+3,(-5,28) is y=-10x-22

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tangent of f(x)=x^2+3,(-5,28)

tangent of

f (x) = x^{2} + 3, (- 5, 28)

y = - 10 x - 22

Solution steps

Find the slope of

f (x) = x^{2} + 3 : \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 = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 28) : y = - 10 x - 22

y = - 10 x - 22

10 y^{^{''}} - 10 y^{^{'}} + 5 y = 0

\int \frac{x ^{3} + 69 x + 84}{x ^{3} + 22 x} d x

\int cos (\frac{nπ x}{4}) d x

\int_{- 3}^{3} ∣ x - 1 ∣ d x

\int (x + 2)^{2} (1 - x)^{5} d x

What is the tangent of f(x)=x^2+3,(-5,28) ?
The tangent of f(x)=x^2+3,(-5,28) is y=-10x-22