Find the point(s) where the line tangent to the graph of f(x)=x^2-4 is parallel to the line y=3x-8
parallel lines have the same slope; so when the derivative = '3' in this case; they would be parallel lines
yup, know that much at least :-)
whats our derivative?
if I take the deriv I have the slope for the tangent and the -reciprocal will give me the slope of the normal line
2x
the normal is perpendicular to the line provided; you want the derivative itself to equal the slope to be parallel so when does 2x = 3?
plug that value for x into the original equation to find the value for y and that will be the point tha tthe tangent line to the curve is parallel to the eqaution of the line that is given
just going through it all - thanks!
so are we looking for a point on f(x) that would have the same slope as the the y (parallel line)?
yes
so 2x=3 when x = 3/2
yes; that will be your x coordinate; now use that to find the y coordinate in the equation that you orginally derived
oh, so that is my x coordinate; that is where I got lost.
did we take the derivative of the y equation to find it's slope at 3? or is that just cause it is the 3 derived from the line;s equation?
we just used the fact that the line equation has a slope of '3' at all points. And we just want to determine a point on our curve that has a parallel slope to it
since the derivative is an equation FOR the slope at any given point; when the derivative = 3, we are parallel to the line given
gotcha, think I am not being clear with my question. how did you find the slope for this line y=3x-8 (i know it seems easy, but I want to see if you used the deriv or another way i.e. cause the 3 is next to the x) Thanks!
it is in form y=mx+b where m is slope or you can take derivative
i just used my extensive knowledge of lines to recall that the 3 sitting next to the x defines the slope of the line ....
we can derive it if we want; [3x-8]' = 3 ; which tells us that the slope at any given point = 3 ...
thank you! that is all very clear now!
:)
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