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?

2. Since it is a quadratic equation, it has the form of a parabola. You know where it intersects the x-axis, and you also know 1 other point, which tells you which way the parabola opens up. From there you can draw a sketch of the graph and then see where it would intersect a line of sloope -4.

3. Because it has root 3 and -5, $y = \alpha (x-3)(x+5) = \alpha (x^2+2x-15)$
$y(1) =-12\alpha=5 \implies \alpha = \frac{-5}{12}$
$y = \frac{-5}{12}(x^2+2x-15) = -\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4}$
Set $-\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4} = -4x+b$ and solve for b. All value smaller than this b will intersect since the parabola is tipped down.

4. Originally Posted by vincisonfire
Because it has root 3 and -5, $y = \alpha (x-3)(x+5) = \alpha (x^2+2x-15)$
$y(1) =-12\alpha=5 \implies \alpha = \frac{-5}{12}$
$y = \frac{-5}{12}(x^2+2x-15) = -\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4}$
Set $-\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4} = -4x+b$ and solve for b. All value smaller than this b will intersect since the parabola is tipped down.
How did you come up with this and how do i solve for b?