Let's work the first problem before moving to the second. Using the point-slope formula, with unknown slope , what is the family of lines which pass through the given point?
1. Find an equation for the straight lines which pass through the point (1, -2) and are tangent to the parabola with equation y = x^{2}.
Would my two equations be y = x^{2} and -2 = m + c? Where do I go from here? Usually I would make equation 1 equal equation 2, but y is not equal to -2.
2. Solve the following for x by completing the square:
2x^{2} - 4x + 1 = 0
What I tried was:
2(x^{2 }- (4/2)x + (1/2)) =0
(x - 1)^{2} + 1 + (1/2) - 1 = 0
(x - 1)^{2 }- (1/2) = 0
x - 1 = sqrt(1/2) = 0
x = 1 + sqrt(1/2)
Thanks in advance.
Yes, this will give you the two lines that satisfying the given conditions:
y=x^2,y=(2+2sqrt(3))(x-1)-2,y=(2-2sqrt(3))(x-1)-2 where x=-2 to 4 - Wolfram|Alpha