# Need help with 2 quadratics problems...

• Oct 14th 2009, 03:42 PM
Andrew00830
Need help with 2 quadratics problems...
Hello, I need some help with 2 quadratics problems:

1. Let f, g be distinct quadratics. In how many places can f and g intersect? Give examples of each case. Is anything else possible? Why?

2. So far we have been finding the vertex of a quadratic f by evaluating h = -b over 2aand k = f(h); find a formula for k in terms of a, b, c.

Thank you *SO* much for anyones help in advance...
• Oct 14th 2009, 04:08 PM
skeeter
Quote:

Originally Posted by Andrew00830
Hello, I need some help with 2 quadratics problems:

1. Let f, g be distinct quadratics. In how many places can f and g intersect? Give examples of each case. Is anything else possible? Why?

sketch two different parabolas in various configurations ... how many ways can you make them intersect (or not)?

2. So far we have been finding the vertex of a quadratic f by evaluating h = -b over 2aand k = f(h); find a formula for k in terms of a, b, c.

$\displaystyle \textcolor{red}{h = \frac{-b}{2a}}$

$\displaystyle \textcolor{red}{f(x) = ax^2 + bx + c}$

find $\displaystyle \textcolor{red}{k = f\left(\frac{-b}{2a}\right)}$

...
• Oct 14th 2009, 04:24 PM
RRH
Quote:

Originally Posted by Andrew00830
Hello, I need some help with 2 quadratics problems:

1. Let f, g be distinct quadratics. In how many places can f and g intersect? Give examples of each case. Is anything else possible? Why?

2. So far we have been finding the vertex of a quadratic f by evaluating h = -b over 2aand k = f(h); find a formula for k in terms of a, b, c.

Thank you *SO* much for anyones help in advance...

If you need help visualizing how your first question looks ... try this ... I know it is sloppy, but it should get the idea across
• Oct 14th 2009, 04:36 PM
Andrew00830
Would question #2 look like this?

k=f(h)=f(-b/(2a))=a*(-b/(2a))^2+b*(-b/(2a))+c=(4ac-b^2)/(4a)
• Oct 14th 2009, 04:54 PM
skeeter
Quote:

Originally Posted by Andrew00830
Would question #2 look like this?

k=f(h)=f(-b/(2a))=a*(-b/(2a))^2+b*(-b/(2a))+c=(4ac-b^2)/(4a)

looks fine to me.