What is the tangent of f(x)=2x^2-9x+10,\at x=1 ?

The tangent of f(x)=2x^2-9x+10,\at x=1 is y=-5x+8

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tangent of f(x)=2x^2-9x+10,\at x=1

tangent of

f (x) = 2 x^{2} - 9 x + 10, at x = 1

y = - 5 x + 8

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = 2 x^{2} - 9 x + 10 : \frac{df ( x )}{d x} = 4 x - 9

Compute the slope of

f (x) = 2 x^{2} - 9 x + 10

x = 1 : m = - 5

Find the line with slope m=

- 5

and passing through

(1, 3) : y = - 5 x + 8

y = - 5 x + 8

y^{^{'}} + \frac{1}{x} y = \frac{2}{3} x^{4} y^{4}

y = x^{2} + 2 x + 11

\frac{d y}{d x} = e (3 x + 2 y)

\frac{d}{d x} (ln (3 x + 2 x^{2}))

\frac{d}{d x} (7 e^{x} csc (x))

What is the tangent of f(x)=2x^2-9x+10,\at x=1 ?
The tangent of f(x)=2x^2-9x+10,\at x=1 is y=-5x+8