What is the tangent of f(x)=x^2-7x+12,\at x=2 ?

The tangent of f(x)=x^2-7x+12,\at x=2 is y=-3x+8

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tangent of f(x)=x^2-7x+12,\at x=2

tangent of

f (x) = x^{2} - 7 x + 12, at x = 2

y = - 3 x + 8

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

f (x) = x^{2} - 7 x + 12 : \frac{df ( x )}{d x} = 2 x - 7

Compute the slope of

f (x) = x^{2} - 7 x + 12

x = 2 : m = - 3

Find the line with slope m=

- 3

and passing through

(2, 2) : y = - 3 x + 8

y = - 3 x + 8

y = 4 - 4 + x

\frac{d}{d r ( t )} (\frac{s - r ( t )}{t})

\frac{\partial}{\partial x} (e - 4 t cos (π x))

\int - cos (x) sin^{2} (x) d x

n = 2 \sum \infty \frac{5}{4 ^{n}}

What is the tangent of f(x)=x^2-7x+12,\at x=2 ?
The tangent of f(x)=x^2-7x+12,\at x=2 is y=-3x+8