What is the tangent of f(x)=x^2+3,(-1,4) ?

The tangent of f(x)=x^2+3,(-1,4) is y=-2x+2

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tangent of f(x)=x^2+3,(-1,4)

tangent of

f (x) = x^{2} + 3, (- 1, 4)

y = - 2 x + 2

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x^{2} + 3

(- 1, 4) : m = - 2

Find the line with slope m=

- 2

and passing through

(- 1, 4) : y = - 2 x + 2

y = - 2 x + 2

\int_{2}^{t} \frac{1}{5 x ^{2}} d x

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

\int \frac{8 x ^{2} - 7 x + 16}{x ^{3} + 4 x} d x

\int_{1}^{3} \frac{1}{x ^{3}} d x

x 3 - x^{2}

What is the tangent of f(x)=x^2+3,(-1,4) ?
The tangent of f(x)=x^2+3,(-1,4) is y=-2x+2