What is the tangent of y=-x^2+4x+2,(3,5) ?

The tangent of y=-x^2+4x+2,(3,5) is y=-2x+11

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tangent of y=-x^2+4x+2,(3,5)

tangent of

y = - x^{2} + 4 x + 2, (3, 5)

y = - 2 x + 11

Solution steps

Find the slope of

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

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

(3, 5) : y = - 2 x + 11

y = - 2 x + 11

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

\frac{\partial}{\partial x} (4 x sin (6 x^{2} y))

\frac{d y}{d x} = x y - x

y^{^{''}} + 2 y^{^{'}} - 3 y = 0

x \to a lim (\frac{1}{x})

What is the tangent of y=-x^2+4x+2,(3,5) ?
The tangent of y=-x^2+4x+2,(3,5) is y=-2x+11