the height of a projectile is modeled by the equation y=-2x^2+38x+10, where x is a time, in seconds, and y is height, in feet, During what interval of time, to the nearest tenth of a second, is the projectile at least 125 feet above ground

y=-2x^2+38x+10

we look for y>125
First we solve for points where y=125:
125=-2x^2+38x+10
-2x^2+38x-115=0
Solving for x, we find
x=3.78 or x=15.22
Thus the projectile is over 125' between t=3.78 s and 15.22 s.

Note: a parabola with the leading coefficient negative (-2x²) is concave downwards.

It's a little confusing.

The original equation for h is
h(x)=-2x^2+38x+10 (Blue line)

I subtracted 125 to find the roots when the projectile exceeds h=125 (red line)
h(x)=-2x^2+38x+10-125, or
h(x)=-2x^2+38x-115

See:
http://img337.imageshack.us/img337/4594/1292008237.png

To find the interval of time during which the projectile is at least 125 feet above the ground, we need to determine the values of x that satisfy the given equation.

The equation y = -2x^2 + 38x + 10 represents the height of the projectile at any given time x. Since we are looking for when the projectile is at least 125 feet above the ground, we can set up the following inequality:

-2x^2 + 38x + 10 ≥ 125

To solve this inequality, we can rearrange it to:

-2x^2 + 38x + 10 - 125 ≥ 0

Combining like terms:

-2x^2 + 38x - 115 ≥ 0

Now, we can factor or use the quadratic formula to solve for x.

Factoring method:

To factor the quadratic equation, we multiply the leading coefficient (-2) by the constant term (-115) to get -2 * -115 = 230.

We need to find two numbers that multiply to 230 and add to the coefficient of the linear term (38).

The two numbers are 23 and 10, as 23 * 10 = 230 and 23 + 10 = 33.

Now, we can rewrite the quadratic equation as:

-2x^2 + 23x + 15x - 115 ≥ 0

Grouping the terms:

(-2x^2 + 23x) + (15x - 115) ≥ 0

Factoring out the common factors:

x(-2x + 23) + 5(3x - 23) ≥ 0

(x - 5)(-2x + 23) ≥ 0

Now we have two factors: (x - 5) and (-2x + 23).

Setting each factor equal to zero and solving for x:

x - 5 = 0 => x = 5

-2x + 23 = 0 => x = 11.5

So, the solutions for x are x = 5 and x = 11.5.

To determine the interval of time, we need to find the values of x between 5 and 11.5 (inclusive), as these are the times when the projectile is at least 125 feet above the ground.

Therefore, the interval of time, to the nearest tenth of a second, during which the projectile is at least 125 feet above ground is [5, 11.5].

graph h vs t.

I don't think it ever gets that high.