9514 1404 393
Answer:
b=-9
intersection point (3, -3)
Step-by-step explanation:
There are a couple of ways to do this. The method that does not involve calculus is to set the two equations equal and find 'b' such that there is exactly one solution.
x^2 -4x = y = 2x +b
In standard form, this is ...
x^2 -6x -b = 0 . . . . . . . we call this the "combined quadratic"
This will have one solution when the discriminant is zero. You may recall that the discriminant of quadratic ax^2 +bx +c is d = (b^2-4ac). We want ...
d = (-6)^2 -4(1)(-b) = 0
36 +4b = 0 . . . . . simplify
b = -36/4 = -9 . . . solve for b
Then the linear equation is ...
y = 2x -9
__
If the discriminant is zero, the solution to the "combined quadratic" is ...
x = -b/(2a) . . . . . . . . with a, b, c from the quadratic form (not y-intercept)
x = -(-6)/(2(1)) = 3
And the point of intersection of the line and the parabola is ...
y = 2(3) -9 = -3 ⇒ (x, y) = (3, -3)
__
The graph is attached.
