What is the tangent of f(x)=2x^2-3x,\at x=4 ?

The tangent of f(x)=2x^2-3x,\at x=4 is y=13x-32

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tangent of f(x)=2x^2-3x,\at x=4

tangent of

f (x) = 2 x^{2} - 3 x, at x = 4

y = 13 x - 32

Solution steps

Find the tangent point:

(4, 20)

Find the slope of

f (x) = 2 x^{2} - 3 x : \frac{df ( x )}{d x} = 4 x - 3

Compute the slope of

f (x) = 2 x^{2} - 3 x

x = 4 : m = 13

Find the line with slope m=

13

and passing through

(4, 20) : y = 13 x - 32

y = 13 x - 32

\frac{d}{d x} (\frac{4 x}{2 - x ^{2}})

\frac{d x}{d t} (1) - x = e^{2 t}

\frac{\partial}{\partial x} (2 x^{2} + y^{2} - 9 y)

2 y^{^{''}} + 7 y^{^{'}} + 5 y = 0

\int_{3}^{5} (- x^{2} + 8 x - 12) d x

What is the tangent of f(x)=2x^2-3x,\at x=4 ?
The tangent of f(x)=2x^2-3x,\at x=4 is y=13x-32