<p>soryy i shouldnt post this question here
but i feel rather desperate. :(
the question is,</p>
<p>for the function f(x)=3x^2 a thte point (4,48), find:</p>
<p>a) the slope of the curve
b) an equation of the normal line</p>
<p>i just dont get what is slope of the 'curve' and what is a 'normal line'</p>
<p>can anyone just tell me how to approach this question?</p>
<p>a) The slope of the curve is the first derivative of the function. After you have the derivative, you plug in x=4 to find the slope of the curve at the point (4, 48).</p>
<p>b) Normal line means perpendicular line. That means the slope for the normal line is the negative reciprocal of the slope you found in part a, and you just plug x=4, y=48, and m=whatever into the point-slope equation.</p>
<p>thanks kelkyann!</p>
<p>Ooh, I'll be in for all that fun this year in calc :D</p>