What is the tangent of f(x)=4x^2+12x+5,\at (-1) ?

The tangent of f(x)=4x^2+12x+5,\at (-1) is y=(8a_{0}+12)x-4a_{0}^2+5

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tangent of f(x)=4x^2+12x+5,\at (-1)

tangent of

f (x) = 4 x^{2} + 12 x + 5, at (- 1)

y = (8 a_{0} + 12) x - 4 a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} + 12 a_{0} + 5)

Find the slope of

f (x) = 4 x^{2} + 12 x + 5 : \frac{df ( x )}{d x} = 8 x + 12

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 = 8 a_{0} + 12

Find the line with slope m=

8 a_{0} + 12

and passing through

(a_{0}, 4 a_{0}^{2} + 12 a_{0} + 5) : y = (8 a_{0} + 12) x - 4 a_{0}^{2} + 5

y = (8 a_{0} + 12) x - 4 a_{0}^{2} + 5

What is the tangent of f(x)=4x^2+12x+5,\at (-1) ?
The tangent of f(x)=4x^2+12x+5,\at (-1) is y=(8a_{0}+12)x-4a_{0}^2+5