If #bbm_1# is the gradient of the normal to a curve at a particular point, and #bbm_2# is the gradient of the tangent at the same point, then:
#m_1*m_2=-1#
Let #m_1# be the gradient of the normal, given:
#m_1=-1/12#
If #m_2# be the gradient of the tangent then:
#m_2=1/(m_1)=-1/(-1/12)=12#
We can find the #bbx# values at the point or points, by using the first derivative, and equating it to #bb12#. Since this function gives us the gradient at a point and we already know this has to be #bb12#
Differentiating
#dy/dx(x^3+3x-1)=3x^2+3#
Equating to 12:
#3x^2+3=12#
#3x^2=9#
#x^2=3=>x=+-sqrt(3)#
Putting these values in #f(x)#, we can obtain two corresponding #y# values:
#f(sqrt(3))=(sqrt(3))^3+3(sqrt(3))-1=6sqrt(3)-1#
#f(-sqrt(3))=(-sqrt(3))^3+3(-sqrt(3))-1=-6sqrt(3)-1#
Using point slope form of a line:
#(y_2-y_1)=m(x_2-x_1)#
For tangent lines we have #m=12#
#y-(6sqrt(3)-1)=12(x-(sqrt(3))#
#color(blue)(y=12x-6sqrt(3)-1\ \ \ \ \ \ \ \ \ \)# tangent line 1
#y-(-6sqrt(3)-1)=12(x-(-sqrt(3))#
#color(blue)(y=12x+6sqrt(3)-1\ \ \ \ \ \ \ \ \ \)# tangent line 2
For normals we have #bb(m=-1/12)#
As above:
#y-(6sqrt(3)-1)=-1/12(x-(sqrt(3))#
#color(blue)(y=-1/12x+73/12sqrt(3)-1\ \ \ \ \ \ \ \)# normal line 1
#y-(-6sqrt(3)-1)=-1/12(x-(-sqrt(3))#
#color(blue)(y=-1/12x-73/12sqrt(3)-1\ \ \ \ \ \ \ \)# normal line 2
GRAPH: