Determine the equations of two lines that pass through the point (-1,-3) and are tangent to the graph of y= x^2+1

Question

this is what i have so far

y'=2x

m = 2(-1)

m = -2

Answer 1

y' = 2x = the slope of the tangent line

but don't plug in the point, (yet), it's not on the parabola

it may help to sketch the graph and plot the point

it's an upward opening parabola with vertex (0,1)

(-1,-3) is below and to the left of the vertex. two lines through the point tangent to the parabola and through (-1,-3) will have a positive slope and the other line a negative slope. one point of tangency will be in quadrant I and the other in quadrant II.

y+3 = 2x(x+1)

x^2 +1 +3 = 2x^2 +2x

x^2 +2x -4 = 0

x = -1 + or - sqr5

y = (-1-sqr5)^2 +1, or (-1+sqr5)^2 +1

x = about -3.2 or 1.2

y = 2.44 or 11.24

tangent points are (-3.2, 2.44) and (1.2, 11.24

tangent lines have slopes = 2x = -2+2sqr5 or -2-2sqr5 = about 2.4 or -6.4

the lines are y =(-2+2sqr5)x + b, solve for b by plugging in (-1,-3)

-3 = (-2 +2sqr5)(-1) + b

b = -3 -2 -2sqr5 = -5 +2sqr5

one tangent line is y = (-2+2sqr5)x -5 +2sqr5

the other line is y = (-2-2sqr5)x + b, find b by plugging in (-1,-3)

-3 = (-2-2sqr5)(-1) + b

b = -3 -2 -2sqr5

the 2nd tangent line is y = (-2-2sqr5)x -5 -2sqr5

or about

y = 2.44x - 0.6

and

y = -6.4x -9.4

(I can't guarantee the arithmetic, but from looking at a rough sketch of the graph and lines, it looks about right. still the answer seems to unnecessarily irrational, maybe there's a mistake or typo in the problem?)