Algebra/Linear Quadratic Systems
Hello could you help me with these questions im unsure how to do them.
Determine the restrictions on the y-intercept so that y = 3x2 + 6x - 1 intersects with a line with slope 2 in more than one place.
A quadratic function has roots 3 and - 5, which passes through the point (1, 5). Under what conditions will it intersect with a line of slope -4 once? Twice? Never?
When we look at where the parabola crosses the y axis, that is when x = 0.
That gives us y = 0 + 0 - 1 = -1. For a line to cross the parabola twice,
when in the form y = mx + b, we need b>-1.
To have roots at 3 and -5, the parabola would have the form f(x) = D(x-3)(x+5).
Since (1,5) is on that curve, we have the equation 5 = D(1-3)(1+5).
Since 1-3 = -2 and 1+5 = 6, that is 5 = D(-2)(6), so D = -5/12.
That gives us the equation f(x) = -5(x-3)(x+5)/12.
That multiplies out to -5(x²+2x-15)/12.
The equation of a line with slope -4 would be y = -4x + b.
This means we would have to be able to solve -4x + b = -5(x²+2x-15)/12.
To get rid of fractions, multiply both sides by 12.
This gives -48x + 12b = -5x² - 10x + 150.
Adding 5x² + 10x - 150 to both sides gives 5x² - 38x - 150 + 12b.
Using the quadratic equation, the square-root in there would have to have a positive value.
The square-root term is b²-4ac, and that is (-38)² - 4*5(-150+2b), which needs to be positive.
Since 38² = 1,444, 4*5*150 = 3,000, and 4*5*2 = 40, that gives 1,444 + 3,000 - 40b.
That gives 4,444 - 40b, and to be positive, we need 40b < 4,444.
Now this is equal at b = 111.1, so as long as b < 111.1, there are two solutions.
At b = 111.1, there is only one solution.