Posted by Elli on Monday, December 19, 2011 at 2:50pm.
A stunt pilot is testing a new plane. The equation that models his height over time is f(x)=15x^2 195x +950, where x is his the time in seconds and (fx) is his height in metres. Determine when the pilot is below 500 metres.
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 3:16pm
Now 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 3:23pm
So do I have factor out the 15x^2  195x + 450, which will give me (x?)(x?)

Math  Steve, Monday, December 19, 2011 at 4:51pm
yes  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 6:00pm
I 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(x10)(x3)
x=10 and x=3
It doesn't seem right~

Math  Steve, Monday, December 19, 2011 at 6:06pm
seems 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 6:35pm
Nevermind
Thank you for clarifications!
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