• Dec 3rd 2008, 06:45 PM
euclid2
A quadratic function has roots $\displaystyle 3$ and $\displaystyle -5$ which passes through the point $\displaystyle (1,5)$. Under what conditions will it intersect with a line of slope $\displaystyle -4$ Once? twice? Never?
• Dec 3rd 2008, 06:50 PM
terr13
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.
• Dec 3rd 2008, 06:51 PM
vincisonfire
Because it has root 3 and -5, $\displaystyle y = \alpha (x-3)(x+5) = \alpha (x^2+2x-15)$
$\displaystyle y(1) =-12\alpha=5 \implies \alpha = \frac{-5}{12}$
$\displaystyle y = \frac{-5}{12}(x^2+2x-15) = -\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4}$
Set $\displaystyle -\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4} = -4x+b$ and solve for b. All value smaller than this b will intersect since the parabola is tipped down.
• Dec 8th 2008, 07:14 PM
euclid2
Quote:

Originally Posted by vincisonfire
Because it has root 3 and -5, $\displaystyle y = \alpha (x-3)(x+5) = \alpha (x^2+2x-15)$
$\displaystyle y(1) =-12\alpha=5 \implies \alpha = \frac{-5}{12}$
$\displaystyle y = \frac{-5}{12}(x^2+2x-15) = -\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4}$
Set $\displaystyle -\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4} = -4x+b$ 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?