What is the tangent of f(x)=x^2-5x+5,\at x=a ?

The tangent of f(x)=x^2-5x+5,\at x=a is y=(2a-5)x-a^2+5

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tangent of f(x)=x^2-5x+5,\at x=a

tangent of

f (x) = x^{2} - 5 x + 5, at x = a

y = (2 a - 5) x - a^{2} + 5

Solution steps

Find the tangent point:

(a, a^{2} - 5 a + 5)

Find the slope of

f (x) = x^{2} - 5 x + 5 : \frac{df ( x )}{d x} = 2 x - 5

Compute the slope of

f (x) = x^{2} - 5 x + 5

x = a : m = 2 a - 5

Find the line with slope m=

2 a - 5

and passing through

(a, a^{2} - 5 a + 5) : y = (2 a - 5) x - a^{2} + 5

y = (2 a - 5) x - a^{2} + 5

\int \frac{- x}{x - 1} d x

\int sin^{3} (θ) cos^{4} (θ) d θ

\int \frac{4}{x ln ^{3} ( x )} d x

e^{- (t - 3)}

x \to 0 + lim (\frac{4}{5 x ^{\frac{1}{3}}})

What is the tangent of f(x)=x^2-5x+5,\at x=a ?
The tangent of f(x)=x^2-5x+5,\at x=a is y=(2a-5)x-a^2+5