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

The tangent of f(x)=2x^2-x-3,\at x=1 is y=3x-5

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

tangent of

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

y = 3 x - 5

Solution steps

Find the tangent point:

(1, - 2)

Find the slope of

f (x) = 2 x^{2} - x - 3 : \frac{df ( x )}{d x} = 4 x - 1

Compute the slope of

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

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, - 2) : y = 3 x - 5

y = 3 x - 5

y = 2 - 6 x

y = \frac{2 x + 4}{1 - x}

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

\frac{d N}{d t} + N = Nt e^{t + 4}

\int x + 11 d x

What is the tangent of f(x)=2x^2-x-3,\at x=1 ?
The tangent of f(x)=2x^2-x-3,\at x=1 is y=3x-5