Question

Find the equation of tangent and normal line to the given curves...

1.) y = x^3 - 7x + 6 at its points of intersection with the x-axis.
...?

Answer 1

First, solve the equation x³-7x+6=0 and we get its points of intersection with x-axis.
Try factors of 6.
x = 1
1³-7×1+6=1-7+6=-6+6=0
x=2
2³-7×2+6=8-14+6=-6+6=0
x=-3
(-3)³-7(-3)+6=-27+21-6=-6+6=0

Then find the derivative of the function y=x³-7x+6.
y'=(x³-7x+6)'=3x² - 7

Find the tangent line to a point using the formula y=f'(a)(x-a)+f(a)
whereas a are the x-intercepts of the graph.

a=1
f(1)=1³-7×1+6=0
f'(1)=3×1² - 7=3-7=-4
y=f'(a)(x-a)+f(a)
y(1)=f'(1)(x-1)+f(1)
y=-4(x-1)+0
y=-4x+4

To find the normal line, just find the line that is perpendicular to the tangent line and passing through that same point (1, 0).


`m=- (1)/(-4) = (1)/(4) \\(1,0) \Rightarrow x_1=1,y_1=0 \\y-y_1=m(x-x_1) \\y-0=(1)/(4)(x- 1) \\y=(1)/(4)x- (1)/(4)`

And similar for other two points: x=2 and x=-3