What is the tangent of y=15-x^2,(1,14) ?

The tangent of y=15-x^2,(1,14) is y=-2x+16

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tangent of y=15-x^2,(1,14)

tangent of

y = 15 - x^{2}, (1, 14)

y = - 2 x + 16

Solution steps

Find the slope of

y = 15 - x^{2} : \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 = - 2

Find the line with slope m=

- 2

and passing through

(1, 14) : y = - 2 x + 16

y = - 2 x + 16

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

y^{^{''}} + 1 = 0

\frac{d}{d y} (3 y)

\int \frac{12 y ^{2}}{4 y ^{3} + 1} d y

\frac{\partial}{\partial x} (sin (8 x - 3 y))

What is the tangent of y=15-x^2,(1,14) ?
The tangent of y=15-x^2,(1,14) is y=-2x+16