What is the tangent of f(x)=x^2+3x+5,(1,9) ?

The tangent of f(x)=x^2+3x+5,(1,9) is y=5x+4

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tangent of f(x)=x^2+3x+5,(1,9)

tangent of

f (x) = x^{2} + 3 x + 5, (1, 9)

y = 5 x + 4

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 5

Find the line with slope m=

5

and passing through

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

y = 5 x + 4

x \to 2 + lim (\frac{1}{∣ 2 - x ∣})

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

\int_{0}^{\frac{π}{6}} sin^{4} (6 x) d x

in v erse l a pl a ce \frac{10}{( s ) ( s ^{2} + 8 s + 25 )}

\int \frac{( x ^{2} + x - 10 )}{( 2 x - 3 ) ( x ^{2} + 4 )} d x

What is the tangent of f(x)=x^2+3x+5,(1,9) ?
The tangent of f(x)=x^2+3x+5,(1,9) is y=5x+4