# Finding the "normal line" of a graph.

• October 4th 2009, 10:51 AM
Rumor
Finding the "normal line" of a graph.
Here's the problem:

"Given the parabola $y = x^2 - Ax + B$, where A is the number of letters in your first name and B is the number of letters in your last name, find an equation for the line L that contains the point (0,B) and is perpendicular to the tangent line to the parabola at (0,B). This is called the 'normal line' to the graph."

For me, A = 8 and B = 6. It says that I can find the slope of the tangent line and then use that to compute the slope of the normal line, but I'm not sure how to go about doing that. I know the derivative of the function is $y = 2x + A$ (or is it $y = 2x - A$?), but I don't know where to go from there.
• October 4th 2009, 01:26 PM
Rumor
Is there no one that can help me with this? D:
• October 4th 2009, 02:05 PM
e^(i*pi)
Quote:

Originally Posted by Rumor
Here's the problem:

"Given the parabola $y = x^2 - Ax + B$, where A is the number of letters in your first name and B is the number of letters in your last name, find an equation for the line L that contains the point (0,B) and is perpendicular to the tangent line to the parabola at (0,B). This is called the 'normal line' to the graph."

For me, A = 8 and B = 6. It says that I can find the slope of the tangent line and then use that to compute the slope of the normal line, but I'm not sure how to go about doing that. I know the derivative of the function is $y = 2x + A$ (or is it $y = 2x - A$?), but I don't know where to go from there.

$\frac{d}{dx}(x^2-Ax+B) = 2x-A$

Compare to the equation of a line: $y = mx+c$

For the normal line recall the product of perpendicular lines is -1