What is the tangent of f(x)=-8x^2-7,\at x=1 ?

The tangent of f(x)=-8x^2-7,\at x=1 is y=-16x+1

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tangent of f(x)=-8x^2-7,\at x=1

tangent of

f (x) = - 8 x^{2} - 7, at x = 1

y = - 16 x + 1

Solution steps

Find the tangent point:

(1, - 15)

Find the slope of

f (x) = - 8 x^{2} - 7 : \frac{df ( x )}{d x} = - 16 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 16

Find the line with slope m=

- 16

and passing through

(1, - 15) : y = - 16 x + 1

y = - 16 x + 1

What is the tangent of f(x)=-8x^2-7,\at x=1 ?
The tangent of f(x)=-8x^2-7,\at x=1 is y=-16x+1