What is the tangent of f(x)=2x^2+7x,\at x=4 ?

The tangent of f(x)=2x^2+7x,\at x=4 is y=23x-32

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tangent of f(x)=2x^2+7x,\at x=4

tangent of

f (x) = 2 x^{2} + 7 x, at x = 4

y = 23 x - 32

Solution steps

Find the tangent point:

(4, 60)

Find the slope of

f (x) = 2 x^{2} + 7 x : \frac{df ( x )}{d x} = 4 x + 7

Compute the slope of

f (x) = 2 x^{2} + 7 x

x = 4 : m = 23

Find the line with slope m=

23

and passing through

(4, 60) : y = 23 x - 32

y = 23 x - 32

\int (x e^{2 x^{2}}) d x

x \to + 1 + lim (\frac{x ^{2}}{x ^{2} - x})

\frac{d}{d a} (sin (a) + cos (a))

\frac{4}{ln ( x )}

What is the tangent of f(x)=2x^2+7x,\at x=4 ?
The tangent of f(x)=2x^2+7x,\at x=4 is y=23x-32