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

To find the equation of the tangent you have to derivate the given curve
to find the normal line you just need to sum 90º to the tangent inclination

Answer 2

d(x^3-7 x+6)/dx = 3x^2-7

Answer 3

first, solve the equation x^3-7x+6=0 and we get its points of intersection with x-axis.
Then find the derivative of the function which is 3x^2 - 7 and you can 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
To find the normal line, just find the line that is perpendicular to the tangent line and passing through that same point (a, 0)

Answer 4

"at its points of intersection with the x-axis." now you need to do y= 0 and solve this:

3x^2-7 = 0
x = +-sqrt(7/3)

Answer 5

you know diogo you are actually solving for points where the derivative of the function is zero.

Answer 6

Thats pretty much it. anhhuyalex completed my thought

Answer 7

can u give the complete solution!

Answer 8

Oh. My bad, it was a mistake. Sorry

I meant to do this:
y = 0
0 = x^3 - 7x + 6

x = -3 or x = 1 or x = 2

Answer 9

equation of tangent line at -3:y=20(x-(-3)+0(only using the formula)=20(x+3)=20x+60
equation of tangent line at 1: y=-4(x-1)=-4x+4
equation of tangent line at 2: y=5(x-2)=5x-10
For the normal lines I hope you can find the line perpendicular to the tangent lines at (-3,0) (1,0) (2,0)

Answer 10

for the normal lines you can simply invert the tangent inclination:

y = mx + b, where m = tangent inclination.
if you want the normal line you just need to do 1/m and thats it.

by using anhhuyalex's values:
tangent at -3 = y = 20x+60 ... normal = (1/20)x+60
tangent at 1 = y = -4x+4 ... normal = -(1/4)x+4
tangent at 2 = y = 5x-10 ... normal = (1/5)x-10

Answer 11

is that it?