Answer to Question #92543 in Calculus for yoko

Question

Find the equations of tangent line and normal lines of parabola y= 2x+ x squared at point (1,3).

Find all the values of x at which the tangent line to the curve y=1/ x + 4 passes through the origin.

Answer 1

Find the equations of tangent line and normal lines of parabola y= 2x+ x squared at point (1,3).

$$y=2x+x^2, \space \space \space (1,3)$$ $$y'=2+2x, \space \space \space y'(1)=2+2*1=4.$$

Tangent line:

$$y=y'_0(x-x_0)+y_0$$

$$y=4(x-1)+3$$

$$y=4x-1.$$

Normal line:

$$y=-\frac{1}{y'_0}(x-x_0)+y_0$$

$$y=-\frac{1}{4}(x-1)+3$$

$$y=\frac{-x+13}{4}.$$

Find all the values of x at which the tangent line to the curve y=1/ x + 4 passes through the origin.

$$y=\frac 1 {x+4}, \space\space y'=-\frac 1 {(x+4)^2}$$

$$y=y'_0(x-x_0)+y_0$$

$$y=y'_0x+(-y'_0x_0+y_0)$$

line passes through the origin => $-y'_0x_0+y_0=0$

$$-\frac 1 {(x_0+4)^2}*(-x_0)+\frac 1 {x_0+4}=0$$

$$\frac {x_0} {(x_0+4)^2}+\frac {x_0+4} {(x_0+4)^2}=0$$

$$\frac {2x_0+4} {(x_0+4)^2}=0$$

$$2x_0+4=0 \space and \space x_0+4\not=0$$

$$x_0=-2.$$

Answer:

1) Tangent line $y=4x-1,$ normal line $y=\frac{-x+13}{4}.$

2) x=-2.