Hi Kinslee! Great question! For this one, we have 2 tangent lines. Now, first question, what are tangent lines? These are lines that intersect at only one point, essentially they just "scrape" the surface of the parabola. The other thing we know is both of these tangent lines intersect at (0,-3).
First thing I would note down is (0,-3) is a y-intercept. So If we were looking at both tangent lines in slope-intercept form, I can form the equatoin y= mx - 3 by plugging in -3 as the y-intercept or "b" value.
Second thing I know is that the slope of all tangent lines come from the first derivative of the line they intersect, so I will now take the derivative of the equation y=x^2+1. This would be dy/dx = 2x via the power rule (bring the 2 in front of the x, subtract the exponent by 1, and all constants, such as +1, goes to 0).
Now that I have my slope equation, I can substitute that in for "m" as that represents the slope in the equation I found in the first step. So that equation now looks like y = 2x*x - 3, simplified it is y = 2x^2 - 3.
Finally, I know the y point has to be the same for this equation and the parabola at at least 1 point since they intersect. So from there, I set this new equation equal to the parabola equation I started with and solve for x. I have that layed out in the steps below
2x^2 - 3 = x^2 + 1
Subtract x^2 on both sides: x^2 - 3 = 1
Subtract 1 on both sides: x^2 - 4 = 0
Factor: (x+2)(x-2) = 0
Find the zeros: x+2 = 0 x-2 = 0
x = -2, x=2
So our x-values would be x=-2,2.
Let me know if that made sense and if you have any questions. This is a tricky question to explain over typing, but hopefully I conveyed everything. Please let me know though if there's anything missing!
