What is the tangent of f(x)=2x^2-5x,(-1,7) ?

The tangent of f(x)=2x^2-5x,(-1,7) is y=-9x-2

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tangent of f(x)=2x^2-5x,(-1,7)

tangent of

f (x) = 2 x^{2} - 5 x, (- 1, 7)

y = - 9 x - 2

Solution steps

Find the slope of

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

Compute the slope of

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

(- 1, 7) : m = - 9

Find the line with slope m=

- 9

and passing through

(- 1, 7) : y = - 9 x - 2

y = - 9 x - 2

\frac{d}{d x} (\frac{x ^{2} + 3}{x ^{2} + 1})

\frac{d}{d x} (a^{2} + b x + c)

\frac{d}{d x} (\frac{2 x + 3}{x + 4})

x \to 4 lim (\frac{x - 4}{x - 4})

y = 5 x + 6

What is the tangent of f(x)=2x^2-5x,(-1,7) ?
The tangent of f(x)=2x^2-5x,(-1,7) is y=-9x-2