Respuesta :
Answer:
The answer to this question can be defined as follows:
Step-by-step explanation:
In point (a) Equation:
[tex]\to y=x^3- 16x + 6[/tex]
point: (4, 6)
Tangent of slop:
[tex]\to m = \frac{dy}{dx}(3x^2-16) \\\\[/tex]
[tex]= (3*16)-16 \\\\ = 48-16 \\\\ =32[/tex]
The line is perpendicular to the tangent then slop is:
[tex]\to -\frac{1}{m}\\\\\to -\frac{1}{32}\\[/tex]
The Formula to the equation of a line:
[tex]\bold{(y-y_1) = m (x-x_1)}[/tex]
[tex]x_1= 4 \ \ \ and \ \ \ y_1= 6 \\\\ \to {(y-6) = -\frac{1}{32} (x-4)}\\\\\to {32y+192 = - x+4}\\\\ \boxed{\to {x+32y=-188}}\\\\[/tex]
In option B:
Slope (m) =[tex]3x^2-16[/tex]
To calculate the smallest slope, differentiate m.
[tex]\to \frac{dm}{dx}= 6x =0\\\\\to m= 0[/tex]
Point on the curve is (x,y) = (0,6)
In option C:
[tex]m= 32\\\\\to 3x^2-16 = 32\\\\\to 3x^2 = 48\\\\\to x^2 = \frac{48}{3}\\\\\to x^2 = 16\\\\ \to x^2 =4^2 \\\\\to x = \pm 4 \\\\\to y = 6,6 \\\\\ Equation \ of \ tangent \ at \ (4,6) is:\\\\\to y-6 = 32(x-4)\\\to y - 6 = 32x - 128\\ \boxed{\to 32x - y = 122}\\\ Equation \ of \ tangent \ at \ (-4,6) is:\\\\y-6 = 32(x+4)\\ \to y - 6 = 32x +128 \\ \boxed{\to 32x - y = -134}\\[/tex]
