What is the normal of f(x)=4x^2+7,(-3,43) ?

The normal of f(x)=4x^2+7,(-3,43) is f(x)= 1/24 x+345/8

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normal of f(x)=4x^2+7,(-3,43)

normal of

f (x) = 4 x^{2} + 7, (- 3, 43)

f (x) = \frac{1}{24} x + \frac{345}{8}

Solution steps

Find the slope of

f (x) = 4 x^{2} + 7 : \frac{df ( x )}{d x} = 8 x

Compute the slope of

f (x) = 4 x^{2} + 7

(- 3, 43) : m = - 24

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{24}

Find the line with slope m=

\frac{1}{24}

and passing through

(- 3, 43) : f (x) = \frac{1}{24} x + \frac{345}{8}

f (x) = \frac{1}{24} x + \frac{345}{8}

81 y^{^{''}} - 16 y = 0

x \to 7 lim (0)

\int_{0}^{5} x^{2} - 5 x d x

(1 + x) y^{^{'}} + 2 y = \frac{sin ( x )}{1 + x}

\int sin (5 θ) sin (θ) d θ

What is the normal of f(x)=4x^2+7,(-3,43) ?
The normal of f(x)=4x^2+7,(-3,43) is f(x)= 1/24 x+345/8