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

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

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

tangent of

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

y = - x + 3

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = x^{2} - 3 x + 4 : \frac{df ( x )}{d x} = 2 x - 3

Compute the slope of

f (x) = x^{2} - 3 x + 4

x = 1 : m = - 1

Find the line with slope m=

- 1

and passing through

(1, 2) : y = - x + 3

y = - x + 3

\frac{d y}{d t} + 2 t y = 2 t e^{- t^{2}}

\frac{\partial}{\partial z} (x y cos (z))

\int x^{6} - 4 x^{4} + x + 1 d x

\int \frac{t ^{5}}{t ^{2} + 10} d t

\frac{d y}{d x} = 0.5 (100 - y)

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