Find the equation of the tangent line to the graph of p(x)=-x-3 at x=1.
-1/2x^2 - 3x
How?
the only way to get -x as a derivative is -1/2x^2 and the only way to get -3 as a derivative is -3x
i dont know the specific rule/law/theorem though
I still don't understand.
do you know how to get a derivative going forward
is this really \[p(x)=-x-3\]?
f'(x) = -x - 3
oooooooooooooh
that is not what it says. it says Find the equation of the tangent line to the graph of p(x)=-x-3 at x=1.
my bad, sorry, im not that far yet in math then lol
and since that is a line itself, the tangent line is the same as that line
So the answer would be -x-3?
i believe the answer is -1 because the derivative of p(x)= -x-3 is -1
-1 is the slope for the line you need a point so plug in x=1 in the original equation and then do point slope
if the question is really the one you asked, then you are asked to find the equation of a line tangent to a line
sorry i was so confused before
a rather silly question, but it is the line itself. that is all
so what about this then...Find the equation of the tangent line \to the graph of \[h(x)=2x^3-4x^2 at x=-2\]
now that is a more reasonable question. 1) take the derivative 2) evaluate the derivative at x = -2 to get the slope 3) evaluate the function at x - -2 4) use the point - slope formula
first find the derivative of the equation - 6x^2 - 8x, that is the slope at any given point, plug in -2 for the x value and that is the slope of the line then, plug in -2 into the original equation, that is the y value of the point (-2,y), use that as your point for point slope
So help me work this out... 1)
and it is 6x^2 -8x, not -6 sorry
6x^2-8x?
yes
then, plug in the -2
okay then 2) 6(-2)^2-8(-2)
you plug in -2 into the orignial equation to get the y-coordinate which is -32. then you find the derivative using the power rule (6x^2-8x) and then plug in x=-2 into the derived equation to get the slope at that particular point. the equation would then be y+32=40(x+2)
24+16
yeah, 40
that is the slope, now plug -2 into the original equation
k
then what?
so now i have -32 and 40
your point to plug in to point slope.... \[y-y _{1} = m(x-x _{1})\] so your point is (-2, -32) and your slope or m is 40
So plug it in for me?
Please
i dont have a calculator, and most teachers let you leave it as point slope so it is \[y+32=40(x+2)\]
perfect...thank you
Join our real-time social learning platform and learn together with your friends!