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

The tangent of f(x)=x^2+ax+2,\at x=1 is y=(2+a)x+1

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tangent of f(x)=x^2+ax+2,\at x=1

tangent of

f (x) = x^{2} + a x + 2, at x = 1

y = (2 + a) x + 1

Solution steps

Find the tangent point:

(1, a + 3)

Find the slope of

f (x) = x^{2} + a x + 2 : \frac{df ( x )}{d x} = 2 x + a

Compute the slope of

f (x) = x^{2} + a x + 2

x = 1 : m = 2 + a

Find the line with slope m=

2 + a

and passing through

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

y = (2 + a) x + 1

\int - 6 ln (x^{2} - 1) d x

\int (x cos (4 x)) d x

\frac{d}{d x} (\frac{2 x + 2 x ^{2}}{( 1 + 2 x ) ^{2}})

\int \frac{1 + y ^{2}}{1 + x ^{2}} d x

\frac{d}{d x} (x e^{- x} - e^{- x})

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