Answer
$y=-2x-1$
$y=6x-9$
Work Step by Step
First find the first derivative of the function:
$f(x)=x^2$
$f'(x)=2x$
Use the slope formula:
$m=\frac{y_2-y_1}{x_2-x_1}$
$m=\frac{x^2+3}{x-1}$
Set $m=f'(x)$ and solve for $x$:
$2x=\frac{x^2+3}{x-1}$
$2x^2-2x=x^2+3$
$x^2-2x-3=0$
$(x-3)(x+1)=0$
$x=-1,3$
Now we can find the slopes of the two tangent lines:
$f'(-1)=2(-1)=-2$
$f'(3)=2(3)=6$
Use point slope form to find the equations:
$y-y_1=m(x-x_1)$
$y+3=-2(x-1)$
$y=-2x-1$
$y+3=6(x-1)$
$y=6x-9$
