1 answer:
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Answer:
Step-by-step explanation:
1. rewrite the equation in standard form:
2. find (h,k), the vertex. the vertex is
3. find the 'focal length' of the parabola - the focal length is the distance between the vertex and the focus. from the vertex we can see that the focal length, p, = 3/2
4. Parabola is symmetric around the y-axis and so the asymptote is a line parallel to the x-axis, a distance p from the y coordinate which is at
. Set up the equation:
5. substitute and solve:
hope this helps, ask me questions if you still don't understand.
Complete Question:
In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t² , 2+t− 5t² , 1 + 2t} to the standard basis C = {1, t, t²}. Then, write t² as a linear combination of the polynomials in B.
Answer:
The change of coordinate matrix is :
U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]
Step-by-step explanation:
Let U = {D, E, F} be any vector with respect to Basis B
U = D [1 − 3t²] + E [2+t− 5t²] + F[1 + 2t]..............(*)
U = [D+2E+F]+ t[E+2F] + t²[-3D-5E]...................(**)
In Matrix form;
The change of coordinate matrix is therefore,
To find D, E, F in (**) such that U = t²
D + 2E + F = 0.................(1)
E + 2F = 0.........................(2)
-3D -5E = 1........................(3)
Substituting eqn (2) into eqn (1 )
D=3F...................................(4)
Substituting equations (2) and (4) into eqn (3)
-9F+10F=1
F = 1
Put the value of F into equations (2) and (4)
E = -2(1) = -2
D = 3(1) = 3
Substituting the values of D, E, and F into (*)
U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]