Post a New Question

Math

posted by on .

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 - ,

    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 - ,

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

  • Math - ,

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

  • Math - ,

    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~

  • Math - ,

    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 - ,

    Nevermind

    Thank you for clarifications!

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question