1. horizontal tan line

Find the equation of the normal line to the graph of the function
f(x) = sec^2(pi x) at the point with x-coordinate 1/4.

Where does f(x) have horizontal tangent lines?

Ok, so I pluged in 1/4 for x and solved for y and got 2. Then how do I do the equation of the line? I don't know how to get the slope for the equation. The answer is in point-slope form.

2. Originally Posted by kwivo
Find the equation of the normal line to the graph of the function
f(x) = sec^2(pi x) at the point with x-coordinate 1/4.
Where does f(x) have horizontal tangent lines?

Ok, so I pluged in 1/4 for x and solved for y and got 2. Then how do I do the equation of the line? I don't know how to get the slope for the equation. The answer is in point-slope form.
the equation of a line is of the form y = mx + b, where m is the slope (given by the derivative) and b is the y-intercept.

if we know the slope and a point the line passes through, we can find the equation of the line by using the point-slope form:

that is, solve $y - y_1 = m(x - x_1)$ for $y$.

where $m$ is the slope, and $(x_1,y_1)$ is a point the line passes through

Note: for the equation of the NORMAL, we find the line that is PERPENDICULAR to the tangent line that touches the graph at the same point the tangent line does

we find the horizontal tangent line by finding the point where the slope is zero, and using the same point-slope form

any questions?

3. I'm sorry. I still don't get it. How do I get the slope for the equation? I understand when you say it's perpendicular, so meaning -reciprical of the tan line. But what is the tan line? I'm confused...

4. [quote=kwivo;74999]I'm sorry. I still don't get it. How do I get the slope for the equation? [quote]i told you, the slope is given by the derivative. find the derivative and plug in what value you were told to plug in.
I understand when you say it's perpendicular, so meaning -reciprical of the tan line.
no. when two lines are perpendicular, their SLOPES are the negative reciprocals of each other. so find the slope of the tangent line, then take its negative reciprocal and that will give you the slope of the normal line. then use the point slope form to find the equation of the line as i told you

do you get it now?

5. I get it now. Thanks.