For what value of b will the line y=-2x+b be tangent to the parabola y=3x^2+4x-1 ? (Show your work - 2 marks)
To find the value of b for which the line is tangent to the parabola, we need to find the point of tangency between the line and the parabola.
Let's start by setting the equation of the line equal to the equation of the parabola:
-2x + b = 3x^2 + 4x - 1
Rearranging the equation, we get:
3x^2 + 6x + (b + 1) = 0
Now, for the line to be tangent to the parabola, this quadratic equation should have a single root with multiplicity 2. In other words, the discriminant should be zero (b^2 - 4ac = 0).
For our quadratic equation, a = 3, b = 6, and c = b + 1. Substituting the values into the discriminant formula, we have:
(6)^2 - 4(3)(b + 1) = 0
36 - 12b - 12 = 0
Simplifying further:
24 - 12b = 0
Rearranging the equation, we find:
12b = 24
b = 24 / 12
b = 2
So, the value of b for which the line y = -2x + b is tangent to the parabola y = 3x^2 + 4x - 1 is b = 2.