We are given the equation of a curve;
[tex]2x^2-4x^{\frac{3}{2}}-\frac{8}{x}-1[/tex]To solve this we begin by taking the derivative of this curve. Note that the slope of this curve is its first derivative.
We now have;
[tex]\begin{gathered} \frac{d}{dx}(2x^2-4x^{\frac{3}{2}}-\frac{8}{x}-1 \\ =4x-6x^{\frac{1}{2}}-\frac{8}{x^2} \end{gathered}[/tex]At this point we should note that the slope (gradient) is the value of this first derivative when x = 4.
We can now plug in this value and we'll have;
[tex]\begin{gathered} f^{\prime}(x)=4x-6x^{\frac{1}{2}}-\frac{8}{x^2} \\ At\text{ } \\ x=4,\text{ we would have;} \\ f^{\prime}(4)=4(4)-6(4)^{\frac{1}{2}}-\frac{8}{4^2} \\ f^{\prime}(4)=16-6(2)-\frac{8}{16} \\ f^{\prime}(4)=16-12-\frac{1}{2} \\ f^{\prime}(4)=3\frac{1}{2} \\ OR \\ f^{\prime}(4)=\frac{7}{2} \end{gathered}[/tex]Now we can see the slope of the curve. The slope of the normal line perpendicular to the tangent of the curve is a negative inverse of this.
The negative inverse of 7/2 would be;
[tex]\begin{gathered} \text{Gradient}=\frac{7}{2} \\ \text{Gradient of perpendicular}=-\frac{2}{7} \end{gathered}[/tex]Now to use this value to derive the equation in the form
[tex]ax+by+c=0[/tex]We start by expresing this in the form;
[tex]y=mx+b[/tex]We now have;
[tex]y=-\frac{2x}{7}+b[/tex]We can convert this to the standard form as indicated earlier;
[tex]\begin{gathered} From\text{ the original equation; when} \\ x=4 \\ y=2(4)^2-4(4)^{\frac{3}{2}}-\frac{8}{4}-1 \\ y=32-4(8)-2-1 \\ y=32-32-2-1 \\ y=-3 \end{gathered}[/tex]With the points
[tex](4,-3)[/tex]We now have, the equation;
[tex]\begin{gathered} y=mx+b \\ -3=-\frac{2(4)}{7}+b \\ -3=-\frac{8}{7}+b \end{gathered}[/tex]We now collect like terms;
[tex]\begin{gathered} b=\frac{8}{7}-3 \\ b=-\frac{13}{7} \end{gathered}[/tex]We now have the y-intercept as calculated above.
We can now write up our equation is the standard form as indicated from the beginning;
[tex]\begin{gathered} ax+by+c=0 \\ (x,y)=(4,-3) \\ c=-\frac{13}{7} \end{gathered}[/tex][tex]\begin{gathered} 4a+(-3)b+(-\frac{13}{7})=0 \\ 4a-3b-\frac{13}{7}=0 \end{gathered}[/tex]Note that A, B and C must be integers. Therefore, we multiply all through by 7;
ANSWER:
[tex]28a-21b-13=0[/tex]