What is the tangent of f(x)=x^2+x,(-4,12) ?

The tangent of f(x)=x^2+x,(-4,12) is y=-7x-16

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tangent of f(x)=x^2+x,(-4,12)

tangent of

f (x) = x^{2} + x, (- 4, 12)

y = - 7 x - 16

Solution steps

Find the slope of

f (x) = x^{2} + x : \frac{df ( x )}{d x} = 2 x + 1

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 = - 7

Find the line with slope m=

- 7

and passing through

(- 4, 12) : y = - 7 x - 16

y = - 7 x - 16

t an g e n t y = 2 x^{3} + 3 x^{2} - 12 x + 5

n = 1 \sum \infty e^{\frac{8}{n}} \cdot \frac{1}{n}

3 \frac{d ^{2} y}{d x ^{2}} - 21 \frac{d y}{d x} + 36 y = 5 sin (3 x)

d er i v a t i v e \frac{1}{x ^{3}}

\frac{\partial}{\partial y} (2 x y^{2} + 3 x^{2})

What is the tangent of f(x)=x^2+x,(-4,12) ?
The tangent of f(x)=x^2+x,(-4,12) is y=-7x-16