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).
[tex]m=- \frac{1}{-4} = \frac{1}{4}
\\(1,0) \Rightarrow x_1=1,y_1=0
\\y-y_1=m(x-x_1)
\\y-0=\frac{1}{4}(x- 1)
\\y=\frac{1}{4}x- \frac{1}{4} [/tex]
And similar for other two points: x=2 and x=-3
