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.

Do i sub in 500 metres for f(x)?

Why did you not follow Steve's suggestion when he answered you the last time you posted this question.

He is right, do what he said.

Yes, to determine when the pilot is below 500 meters, you need to substitute 500 for f(x) in the equation.

So, the equation becomes:

500 = 15x^2 - 195x + 950

Now, you need to rearrange the equation to bring it to the standard quadratic form, which is ax^2 + bx + c = 0.

Subtracting 500 from both sides:

15x^2 - 195x + 450 = 0

At this point, you have a quadratic equation that can be solved using various methods, such as factoring, completing the square, or using the quadratic formula. Given the coefficients of the equation, factoring is the most suitable method in this case.

To factor the equation, you can begin by looking for two numbers that multiply to give you the product of the coefficient of x^2 term (15) and the constant term (450), and add up to the coefficient of the x term (-195). In this case, the numbers are -15 and -30.

So, you can rewrite the equation as:

15x^2 - 15x - 30x + 450 = 0

Now, you can factor by grouping:

(15x^2 - 15x) + (-30x + 450) = 0

Taking out the common factor from the first two terms and the last two terms separately:

15x(x - 1) - 30(x - 15) = 0

Now, you can factor out (x - 1) from the first two terms and -30 from the last two terms:

15x(x - 1) - 30(x - 15) = 15x(x - 1) - 30(x - 1)

This simplifies to:

(15x - 30)(x - 1) = 0

Now, set each factor equal to zero:

15x - 30 = 0 or x - 1 = 0

Solve each equation separately:

15x - 30 = 0
15x = 30
x = 2

x - 1 = 0
x = 1

So, the pilot is below 500 meters at two different times: when x = 1 second and when x = 2 seconds.