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.

Question

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?

Answer 1

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.

Answer 2

So do I have factor out the 15x^2 - 195x + 450, which will give me (x-?)(x-?)

Answer 3

yes - just solve for x the way you normally do, either by grouping as you say, or by the quadratic formula

Answer 4

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(x-10)(x-3)

x=10 and x=3

It doesn't seem right~

Answer 5

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.

Answer 6

Nevermind

Thank you for clarifications!