What is the tangent of-x^2+3x+4,\at x=1 ?

The tangent of-x^2+3x+4,\at x=1 is y=x+5

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tangent of-x^2+3x+4,\at x=1

tangent of

- x^{2} + 3 x + 4, at x = 1

y = x + 5

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

y = - x^{2} + 3 x + 4 : \frac{d y}{d x} = - 2 x + 3

Compute the slope of

y = - x^{2} + 3 x + 4

x = 1 : m = 1

Find the line with slope m=

1

and passing through

(1, 6) : y = x + 5

y = x + 5

\frac{d y}{d x} = \frac{- e ^{x}}{8}

\frac{\partial}{\partial y} (x y e^{z})

cos (7 x)

x \to - 1 - lim ({2 x^{2} - 5 x})

\frac{d}{d x} (3 sec^{2} (π x - 1))

What is the tangent of-x^2+3x+4,\at x=1 ?
The tangent of-x^2+3x+4,\at x=1 is y=x+5