Find the equation of the normal to the curve y=3x²-2x-1 which is parallel to the line y=x-3.
the slope of the line y = x - 3 is 1. thus we want this to be the slope of our normal line. however, the normal line is perpendicular to the tangent line, so the slope of the tangent line at the said point is -1. since it is easier to find the slope of the tangent line, we will do so.
- differentiate the function
- set the derivative equal to -1 and solve for x. this will give you the x-coordinate of the point you want. use it to find the y-coordinate
- when you have the point, use the point slope form:
$\displaystyle y - y_1 = m(x - x_1)$ where $\displaystyle m = 1$ and $\displaystyle (x_1,y_1)$ is the point i asked you to find
- solve for y to get the equation of the line
- post your solution so we can check
y=x-3
Gradient (m) = 1
Gradient of normal = -1/1 = -1
y=3x²-2x-1
dy/dx = 3*2*x²-¹ - 2*1*x¹-¹ = 6x-2
6x-2 = -1
6x = -1+2
6x = 1
x = 6/1
y = 3x²-2x-1
y = 3(1/6)²-2(1/6)-1
y = -5/4
y-y1 = m(x-x1)
y--5/4 = -1(x-1/6)
y+5/4 = -x +1/6
y = -x +1/6-5/4
y = -x -13/12
The answer writen in the text book is 12y = 12x -17
$\displaystyle y=x-3$
$\displaystyle \implies Gradient~of~normal = 1$
$\displaystyle \implies Gradient~of~tangent = -1$
$\displaystyle y=3x^2-2x-1$
$\displaystyle \implies \frac{dy}{dx} = 6x-2$
$\displaystyle \frac{dy}{dx} = -1 \implies 6x-2 = -1$
$\displaystyle \implies x = \frac {1}{6}$
$\displaystyle x = \frac {1}{6} \implies y = - \frac {5}{4}$
$\displaystyle \implies Equation~of~Normal \rightarrow y-y_{1}=m(x-x_{1})$
$\displaystyle \implies y+\frac {5}{4} = 1(x- \frac {1}{6})$
$\displaystyle \implies y=x- \frac {17}{12}$
$\displaystyle \implies 12y = 12x-17$
When you went to the equation of a straight line you used the gradient for the tangent, which is -1 hence you found the equation of the tangent. Instead you need to use the gradient of the normal, which is 1 and then you would have followed through to the right answer.