# Thread: Need some maths help. I am stuck!

1. ## Need some maths help. I am stuck!

A curve has the equation y=x^2-3x+4

a) Find an equation of the normal to the curve at the point A (2,2).

The normal to the curve at A intersects the curve again at the point B.

b) Find the coordinates of the point B.

Any one know how to do this?

2. Originally Posted by Colin_m
A curve has the equation y=x^2-3x+4

a) Find an equation of the normal to the curve at the point A (2,2).

The normal to the curve at A intersects the curve again at the point B.

b) Find the coordinates of the point B.

Any one know how to do this?
hint: the normal line is the line perpendicular to the tangent line.

do you know how to find the tangent line?

3. do i need to find the gradient then use y=mx+c?

4. Originally Posted by Colin_m
do i need to find the gradient then use y=mx+c?
Yes, but what is the relationship between the slopes of two perpendicular lines?

-Dan

5. there the same

6. Originally Posted by Colin_m
there the same
No, if you have two (non-vertical) perpendicular lines their slope products is -1.

7. i am confused now. So how will i start off working this out?

A curve has the equation y=x^2-3x+4

a) Find an equation of the normal to the curve at the point A (2,2).

The normal to the curve at A intersects the curve again at the point B.

b) Find the coordinates of the point B.

Any one know how to do this?

8. Originally Posted by Colin_m
i am confused now. So how will i start off working this out?

A curve has the equation y=x^2-3x+4

a) Find an equation of the normal to the curve at the point A (2,2).

The normal to the curve at A intersects the curve again at the point B.

b) Find the coordinates of the point B.

Any one know how to do this?
Find your gradient for the function at x = 2. That's the slope of the line tangent to your curve at x = 2. To find the slope of the line normal to this you take the slope of your normal line as -1/gradient.

-Dan

9. by the way, the formula for the slope at any value of x is given by the derivative