00:02
In this question, we are given the expression y is equal to 1 over x squared minus x plus 3 to the power of 3 over 2.
00:17
And we are to find the equation to the equation of the tangent to this curve at the point x is equal to minus 2.
00:28
Now we can rewrite y as x squared minus x plus 3 to the power of minus 3 over 2 and the derivative y prime will be equal to minus 3 over 2 x squared minus x plus 3 to the new power of old power minus 1 so we get minus 5 over 2 multiply by the derivative of the algebraic expression in the brackets.
01:02
We have 2x minus 1.
01:06
Now, plugging in our point, so y prime at x is equal to minus 2, we are going to get minus 3 over 2 bracket, minus 2 squared, that's 4, minus 2, minus 1, minus 2, we get plus 2 and plus 3, to the power of minus 5 over 2, multiply by 2 by minus 2, that's minus 4, minus 1, we get minus 5.
01:37
So here we have minus 5 by minus 3, we get 15 over 2, bracket, 4 plus 2, that's 6, 7, 8, 9.
01:48
So we have 9 to the power of minus 5 over 2.
01:52
And here we get the root of 9 is 3, 3 to the power of minus 5...