The slope of y' =1∴ x ∴ m is the slope of k =1∴ x. ∴ n, k' =-∴ x. and the crossing point of n, p.

∴n: y-y .=-√x .×(x-x .)

∴Q(x。 +(y ./√x .),0)

∴KQ=y。 /√ X. and p is on y = √ X.

∴y。 /√x .= 1

17 and \ are even functions ∴ the odd power term is 0, and f (x) = ax4+cx2+e.

* intersection A(0,-1)∴e=- 1

y'=4ax^3+2cx^2

Y'=4a+2c y=a+c- 1 at x= 1

∴ y-(a+c-1) = (4a+2c) * (x-1) is simplified to y-(4a+2c)*x+3a+c+ 1=0.

∫2x+y-2 = 0∴4a+2c =-23a+c+ 1 =-2，a=-2，c=3。

∴f(x)=-2x^4+3x^2- 1

18, let p. (x X., Y.)

y'=3x？ + 1 k=3x .？ Tangent at+1 ∵P0 L 1 parallel straight line 4x-y- 1=0∴k=4.

∴3x。 ? +1=4 gives x. = 1∶P0 in the third quadrant ∴ X. =-1∴ P. (- 1,-4)

The slope of l is-1/k=- 1/4.

∴L:y+4=- 1/4(x+ 1)