What is the tangent of y=x^2+1,(-4,17) ?

The tangent of y=x^2+1,(-4,17) is y=-8x-15

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tangent of y=x^2+1,(-4,17)

tangent of

y = x^{2} + 1, (- 4, 17)

y = - 8 x - 15

Solution steps

Find the slope of

y = x^{2} + 1 : \frac{d y}{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 = - 8

Find the line with slope m=

- 8

and passing through

(- 4, 17) : y = - 8 x - 15

y = - 8 x - 15

\frac{\partial}{\partial x} (\frac{2}{3 x})

\int \frac{( 3 x ^{3} + 4 x )}{( x ^{2} + 5 ) ^{2}} d x

((x^{2} + 1)^{2})^{^{'}}

y^{^{''}} + 4 y^{^{'}} + 4 y = sec (t) tan (t)

\int_{0}^{1} (x^{2} - x) d x

What is the tangent of y=x^2+1,(-4,17) ?
The tangent of y=x^2+1,(-4,17) is y=-8x-15