You need to find points where the line g(x) intercepts the quadratic function f(x) in one and only one point.
Then 3x^2 + 4x -2 = mx - 5
solve 3x^2 + 4x - mx -2 + 5 = 0
3x^2 + (4 - m)x + 3 = 0
In order to there be only one solution (one intersection point) the radicand of the quadratic formula must be 0 =>
b^2 - 4ac = (4 - m)^2 - 4(3)(3) = 0
(4 - m)^2 = 24
4 - m = +/- √(24)
m = 4 +/- √(24) = 4 +/- 2√(6)
Then m, the slope of the line, may be 4 + 2√6 and 4 - 2√6
Answer:
Extraneous Solutions An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation. Example 1: Solve for x, 1 x − 2 + 1 x + 2 = 4 (x − 2) (x + 2).
Answer:
look up the paper because i had that same paper
Step-by-step explanation:
There are 2 answers for this in my opinion
1. nothing happens because its still in the origin and that won't change
2. depending on which way you turn it the points will change based on that
so if you know how to use a graph you should be good
5(4x - 10) + 10x = 4(2x - 3) + 2(x - 4)
20x - 50 + 10x = 8x - 12 + 2x - 8
30x - 50 = 10x - 20
30x - 10x = -20 + 50
20x = 30
x = 3/2 <==
