What is the tangent of f(x)=x^2-7x+7,(6,1) ?

The tangent of f(x)=x^2-7x+7,(6,1) is y=5x-29

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tangent of f(x)=x^2-7x+7,(6,1)

tangent of

f (x) = x^{2} - 7 x + 7, (6, 1)

y = 5 x - 29

Solution steps

Find the slope of

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

Compute the slope of

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

(6, 1) : m = 5

Find the line with slope m=

5

and passing through

(6, 1) : y = 5 x - 29

y = 5 x - 29

y^{^{''}} + 3 y^{^{'}} + 2 y = 20 e^{3 t}

\frac{3}{s ( s ^{3} + 2 s ^{2} + 2 s + 1 )}

\frac{\partial}{\partial y} (e^{\frac{y}{x}} - \frac{e ^{\frac{y}{x}} y}{x})

t + 3 t

n = 0 \sum \infty 2 (\frac{2}{3})^{3 n}

What is the tangent of f(x)=x^2-7x+7,(6,1) ?
The tangent of f(x)=x^2-7x+7,(6,1) is y=5x-29