What is the tangent of y=x^2+7,(3,16) ?

The tangent of y=x^2+7,(3,16) is y=6x-2

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tangent of y=x^2+7,(3,16)

tangent of

y = x^{2} + 7, (3, 16)

y = 6 x - 2

Solution steps

Find the slope of

y = x^{2} + 7 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} + 7

(3, 16) : m = 6

Find the line with slope m=

6

and passing through

(3, 16) : y = 6 x - 2

y = 6 x - 2

\frac{d}{d x} (x^{3} + 2 x + 1)

\int sin (3 x + 2) d x

y = 4 x cos (x), at x = π

\int \frac{1}{csc ^{2} ( x )} d x

(2.1) (8.7)