Question

What is the equation of the line normal to `f(x)=4x^2-3x-2` at `x=3`?

Answer 1

`y=1/7 x+24 6/7`

Explanation:

Given:`" "f(x)=4x^2-3x-2`.......................Equation(1)

`=>f^'(x)=8x-3`

So the gradient of the normal will be `1/f^'(x)`

Thus we have `y=1/f^'(x) x+c`...........................Equation(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the point on the curve at "x=3)`

`f(3)=4(3)^2-3(3)-2 =25`

Thus for the point `(x,y)->(3,25)` equation(2) becomes:

`25=1/(8(3)-3) (3)+c`

`c=25-1/7 = 24 6/7`

By substitution Equation(2) becomes:

`y=1/7 x+24 6/7`