What is the tangent of f(x)=5x^2+8,(-3,53) ?

The tangent of f(x)=5x^2+8,(-3,53) is y=-30x-37

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tangent of f(x)=5x^2+8,(-3,53)

tangent of

f (x) = 5 x^{2} + 8, (- 3, 53)

y = - 30 x - 37

Solution steps

Find the slope of

f (x) = 5 x^{2} + 8 : \frac{df ( x )}{d x} = 10 x

Compute the slope of

f (x) = 5 x^{2} + 8

(- 3, 53) : m = - 30

Find the line with slope m=

- 30

and passing through

(- 3, 53) : y = - 30 x - 37

y = - 30 x - 37

\frac{d}{d x} ((a + b - 5) x^{2} + (b + c - 7) x + (a + c))

f (x) = x + \frac{3}{x}, at x = 3

(- 5.6) (- 2.5)

\frac{2}{7}

\frac{d}{d x} (tan (x) + cos^{2} (x) + sin^{2} (x) - 3)

What is the tangent of f(x)=5x^2+8,(-3,53) ?
The tangent of f(x)=5x^2+8,(-3,53) is y=-30x-37