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July 24, 2014

Posted by **Elli** on Monday, December 19, 2011 at 2:50pm.

I subbed in 500 for f(x)

f(x)=15x^2 -195x +950

500 = 15x^2 -195x + 950

0=15x^2 -195x + 950 - 500

0=15x^2 - 195x + 450

What do i do next?

- Math -
**Steve**, Monday, December 19, 2011 at 3:16pmNow you can solve for x to find when the pilot was at 500m.

Think of the graph. It's a parabola, opening up. So, the vertex is between the two roots, and below the line y=500.

So, the height will be below 500 for all x between the roots of your equation as shown.

- Math -
**Elli**, Monday, December 19, 2011 at 3:23pmSo do I have factor out the 15x^2 - 195x + 450, which will give me (x-?)(x-?)

- Math -
**Steve**, Monday, December 19, 2011 at 4:51pmyes - just solve for x the way you normally do, either by grouping as you say, or by the quadratic formula

- Math -
**Elli**, Monday, December 19, 2011 at 6:00pmI did factoring and got this:

f(x)=15x^2 -195x +950

500 = 15x^2 -195x + 950

0=15x^2 -195x + 950 - 500

0=15x^2 - 195x + 450

0=15(x^2 - 13x + 30)

0=15(x-10)(x-3)

x=10 and x=3

It doesn't seem right~

- Math -
**Steve**, Monday, December 19, 2011 at 6:06pmseems ok to me. From t=3 to t=10 the plane is below 500 m.

at the vertex, t=13/2, f(6.5)= 316.25

f(2.9) = f(10.1) = 510.65

what bothered you about the answer? you expected to get two values for x.

- Math -
**Elli**, Monday, December 19, 2011 at 6:35pmNevermind

Thank you for clarifications!

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