TANGENTS AND NORMALS

TANGENTS AND NORMALS


C
Asked by 2 months ago
10 million points

Find the equation of the normal to the curve $y=(x^2+x+1)(x-3)$ at the point where it cuts the $x-axis$

TANGENTS AND
chegendungu

1 Answer

C
Answered by 2 months ago
10 million points

$y=(x^2+x+1)(x-3)$

When $y=0$

$(x^2+x+1)(x-3)=0$

But $x^2+x+1=0$ has no real roots,

$\therefore x=+3$

$\therefore$ the curve cuts the $x-axis$ at $(3,0)$

$y=x^3-2x^2-2x-3$

$\therefore$ grad $t=3x^2-4x-2$

When $x=3$

grad $y=27-12-2=13$

The gradient of the tangent at $(3,0)$ is $+13$, therefore the gradient of the normal at

$(3,0)$ is $-\frac{1}{13}$ and its equation is

$\frac{y-0}{x-3}=-\frac{1}{13}$

$\therefore 13y=-x+3$

$\therefore$ the equation of the normal is $x+13y-3=0$

Your Answer

Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$

Use LaTeX to type formulas and markdown to format text. See example.

Sign up or Log in

  • Answer the question above my logging into the following networks
Sign in
Sign in
Sign in

Post as a guest

  • Your email will not be shared or posted anywhere on our site
  •  

Stats
Views: 4
Asked: 2 months ago

Related