Question

What is the equation of the line through (−2, −2) and (4, −5)? A.y = 12 1 2 x – 1 B.y = –12 1 2 x – 3 C.y = 2x + 2 D.y = –2x – 6

Answer 1

y=7x−5

Step-by-step explanation:

We have; y=x4+2x2−x

First we differentiate wrt x;y=x4+2x2−x

∴dydx=4x3+4x-1

We now find the vale of the derivative at (1,2) (and it always worth a quick check to see that y=2 when x=1) we have dydx=4+4−1=7

So at the tangent passes through the coordinate (1,2) and has gradient m=7

We now use y−y1=m(x−x1) to get the equation of the tangent:

∴y−2=7(x−1)

∴y−2=7x−7

∴y=7x−5