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

The tangent of y=2x^2+5,(-1,7) is y=-4x+3

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

tangent of

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

y = - 4 x + 3

Solution steps

Find the slope of

y = 2 x^{2} + 5 : \frac{d y}{d x} = 4 x

Compute the slope of

y = 2 x^{2} + 5

(- 1, 7) : m = - 4

Find the line with slope m=

- 4

and passing through

(- 1, 7) : y = - 4 x + 3

y = - 4 x + 3

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

\int tan^{4} (5 t) d t

\int \frac{1 + 3 x}{( 1 - x ) ( 3 x - 5 )} d x

\int e^{7 t} d t

y = 5 + 5 x^{2} - 2 x^{3}, (2, 9)

What is the tangent of y=2x^2+5,(-1,7) ?
The tangent of y=2x^2+5,(-1,7) is y=-4x+3