What is the tangent of f(x)=x^2+6,(5,31) ?

The tangent of f(x)=x^2+6,(5,31) is y=10x-19

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tangent of f(x)=x^2+6,(5,31)

tangent of

f (x) = x^{2} + 6, (5, 31)

y = 10 x - 19

Solution steps

Find the slope of

f (x) = x^{2} + 6 : \frac{df ( x )}{d x} = 2 x

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 = 10

Find the line with slope m=

10

and passing through

(5, 31) : y = 10 x - 19

y = 10 x - 19

What is the tangent of f(x)=x^2+6,(5,31) ?
The tangent of f(x)=x^2+6,(5,31) is y=10x-19