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

The tangent of f(x)=x^2+x,(-3,1) is y=(2sqrt(5)-5)x+6sqrt(5)-14 \lor y=(-2sqrt(5)-5)x-14-6sqrt(5)

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

tangent of

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

y = (25 - 5) x + 65 - 14 or y = (- 25 - 5) x - 14 - 65

Solution steps

Slope of the tangent line at

(- 3, 1) : m = \frac{x ^{2} + x - 1}{x + 3}

Find the slope of

f (x) = x^{2} + x : m = 2 x + 1

Solve

2 x + 1 = \frac{x ^{2} + x - 1}{x + 3} : x = - 3 + 5, x = - 3 - 5

Find the tangent line at

x = - 3 + 5 : y = (25 - 5) x + 65 - 14

Find the tangent line at

x = - 3 - 5 : y = (- 25 - 5) x - 14 - 65

y = (25 - 5) x + 65 - 14 or y = (- 25 - 5) x - 14 - 65

f (x) = \frac{e ^{x}}{4 x ^{2} + 1}

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

\int \frac{1}{4 + 9 x ^{2}} d x

\frac{d y}{d x} - 4 y = x^{2} + 9

y = 8 x - x^{2} (1.7)

What is the tangent of f(x)=x^2+x,(-3,1) ?
The tangent of f(x)=x^2+x,(-3,1) is y=(2sqrt(5)-5)x+6sqrt(5)-14 \lor y=(-2sqrt(5)-5)x-14-6sqrt(5)