What is the tangent of f(x)=4x^2+3,\at x=1 ?

The tangent of f(x)=4x^2+3,\at x=1 is y=8x-1

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tangent of f(x)=4x^2+3,\at x=1

tangent of

f (x) = 4 x^{2} + 3, at x = 1

y = 8 x - 1

Solution steps

Find the tangent point:

(1, 7)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 8

Find the line with slope m=

8

and passing through

(1, 7) : y = 8 x - 1

y = 8 x - 1

\frac{2}{3}

y = - 3

y = x^{3} - 5 x, (- 1, 4)

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

f (x) = 4 x^{2}

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