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

The tangent of f(x)=2x^2-5x+2,\at x=a is y=(4a-5)x-2a^2+2

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

tangent of

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

y = (4 a - 5) x - 2 a^{2} + 2

Solution steps

Find the tangent point:

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

Find the slope of

f (x) = 2 x^{2} - 5 x + 2 : \frac{df ( x )}{d x} = 4 x - 5

Compute the slope of

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

x = a : m = 4 a - 5

Find the line with slope m=

4 a - 5

and passing through

(a, 2 a^{2} - 5 a + 2) : y = (4 a - 5) x - 2 a^{2} + 2

y = (4 a - 5) x - 2 a^{2} + 2

x^{4} + 5 y^{4} \frac{x ^{5} + 4 d y}{d x} = 0, y (0) = 1

\frac{d y}{d x} - 2 x y = 0

\frac{( x - 10 )}{5 x + 7}

f (x) = \frac{4}{x}, (- 2, - 2)

\frac{d y}{d x} = 1 + e^{y - x + 3}

What is the tangent of f(x)=2x^2-5x+2,\at x=a ?
The tangent of f(x)=2x^2-5x+2,\at x=a is y=(4a-5)x-2a^2+2