Finding the tangent and normals to a quadratic

Question

How can you find the tangent and normals to the equation $y=x^2-7x+6$ at its point of intersection with the $x$-axis.

Answer 1

Roots $(y=0)$ are at $ x= 1 $ and $x=6 $;

Slope = differential coefficient = $ 2 x-7$ which are $(-5,5)$

$$\dfrac{y-y_1}{x-x_1}= 2 * -5 $$

$$\dfrac{y-y_2}{x-x_2}= 2 * +5 $$

Fill up $(x,y)$ coordinates from first line for tangents.

You can likewise do for normals (whose slopes are negative reciprocals of tangents).

Answer 2

1. Find the two intersection points with the x axis
2. For the tangent: Find the derivative and fill in at both points (it gives you the slope)
3. The normal is orthogonal to the tangent (slope is -1/tangent's slope)

For more hints, if necessary, provide more effort from your side.

Answer 3

$\ y=x^2-7x+6 $

$\frac{dy}{dx}=2x-7 $

we know the point to be $\ 0$. Therefore,

slope= $\ -5 $

and,

$\ y=6$

From $\ y=mx+c$

We get: $\ y=c$

$\ c=6$

Therefore, plugging into the following equation $\ y=mx+c$ we get:

$\ y=-5x+6$ is the required equation of tangent

I guess u will be able to find the equation of normal yourself.

Best of Luck!