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

The tangent of f(x)=2x^2+5x-3,\at x=1 is y=9x-5

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

tangent of

f (x) = 2 x^{2} + 5 x - 3, at x = 1

y = 9 x - 5

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = 2 x^{2} + 5 x - 3 : \frac{df ( x )}{d x} = 4 x + 5

Compute the slope of

f (x) = 2 x^{2} + 5 x - 3

x = 1 : m = 9

Find the line with slope m=

9

and passing through

(1, 4) : y = 9 x - 5

y = 9 x - 5

\int \frac{sin ( 2 x )}{41 + cos ^{2} ( x )} d x

\int_{1}^{3} x ln (x) d x

\frac{\partial}{\partial y} (π x x^{2} + y^{2})

\frac{- 1}{( 2 x + 1 ) 2 x + 1}

\frac{d}{d x} (1 - 3 x)

What is the tangent of f(x)=2x^2+5x-3,\at x=1 ?
The tangent of f(x)=2x^2+5x-3,\at x=1 is y=9x-5