<p>can someone help me do this problem...</p>
<p>A curve C is defined by the parametric equations x=t^2-4t+1 and y=T^3. Which of hte following is an equation of the line tangent to the graph of C at point (-3,8)</p>
<p>A)x=-3
B)x=2
C)y=8
D)y=-27/10(x+3)+8
E)12(x+3)+8</p>
<p>explanation too thanks =)</p>
<p>Well the first thing we want to do here, since we’re finding the equation of a line in terms of y and x, is find dy/dx. </p>
<p>dy/dx = (dy/dt)/(dx/dt). So to find dy/dx we first find dy/dt = 3t^2, then dx/dt = 2t - 4.</p>
<p>Thus dy/dx = (3t^2)/(2t-4). But the point provided is in (x,y)! Therefore, we’ll go back to the original equations for x and y to find the proper t value. When x = -3, t^2 - 4t + 1 = -3.This is equivalent to (t-2)^2 = 0, so t = 2. Check with y; is 2^3 = 8? Yes! So t = 2.</p>
<p>Plug that back into dy/dx to get dy/dx = 12/0. This means slope is undefined, and for the point (-3,8) the answer is A, x = -3. (Think about it; both A and B are straight vertical lines, so they have undefined slope, just like dy/dx. However, only x = -3 is the tangent line for (-3,8), since the other point is at a different x-coordinate.)</p>
<p>Answer is A
for the first equation, plug in -3 for x
-3 = t^2-4t+1
0=t^2-4t+4
0=(t-2)^2
t=2
and then 8 for y
8=t^3
t=2</p>
<p>and then find the derivative by dy/dt over dx/dt which is (3t^2)/(2t-4)
plug your two in…
= 12/0
which represents a vertical asymptote. (which is x= a number)</p>
<p>so plug your 2 back into your x equation; 2^2-4(2)+1 = -3</p>
<p>i’m 99% sure it’s A.</p>