What is the tangent of f(x)=-8x^4+11x^2-3,(1,0) ?

The tangent of f(x)=-8x^4+11x^2-3,(1,0) is y=-10x+10

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tangent of f(x)=-8x^4+11x^2-3,(1,0)

tangent of

f (x) = - 8 x^{4} + 11 x^{2} - 3, (1, 0)

y = - 10 x + 10

Solution steps

Find the slope of

f (x) = - 8 x^{4} + 11 x^{2} - 3 : \frac{df ( x )}{d x} = - 32 x^{3} + 22 x

Compute the slope of

f (x) = - 8 x^{4} + 11 x^{2} - 3

(1, 0) : m = - 10

Find the line with slope m=

- 10

and passing through

(1, 0) : y = - 10 x + 10

y = - 10 x + 10

\int \frac{2}{x ^{2} + 3 x} d x

\int \frac{( sin ( x ) )}{cos ( x )} d x

\int \frac{x ^{\frac{1}{3}} - x ^{\frac{1}{4}}}{4 x ^{\frac{1}{2}}} d x

\int \frac{1}{2 + 2 cos ( x )} d x

\frac{d y}{d x} = x e^{x^{2} - y}

What is the tangent of f(x)=-8x^4+11x^2-3,(1,0) ?
The tangent of f(x)=-8x^4+11x^2-3,(1,0) is y=-10x+10