Deon throws a ball into the air. The height h of the ball, in meters, at time t seconds is modeled by the function h(t)=-5t^2+2t+39
Will the ball reach height of 41 meters. What is the discriminate?

Please help me, please

Answer :

Answer:

The ball will not reach a height of 41 meters.

The discriminate is 784

Step-by-step explanation:

You know that the height h of the ball, in meters, at time t seconds is modeled by the function h(t) = - 5*t² + 2*t + 39

To find out if the ball will reach a height of 41 meters, you must calculate the maximum of the function, that is, the highest or highest point on the graph.  That is, the maximum in a function f is the largest value that the function takes.

The vertex is a point that is part of the parabola, which has as its ordinate the maximum value (as in this case) or minimum value of the function. That is, the vertex is the maximum of the function (as in this case) or minimum.

In this case, the maximum of t is obtained by:

[tex]t=\frac{-b}{2*a}[/tex]

Being a=-5 and b=2 (comparing with a quadratic function of the form: f(x)=a*x² + b*x +c)

[tex]t=\frac{-2}{2*(-5)}[/tex]

t= 0.2

The maximum height value must be obtained by substituting the previously calculated value of t in the function:

h(0.2) = - 5*0.2² + 2*0.2 + 39

h(0.2)=39.2

So the maximum height the ball will reach is 39.2 meters. The ball will not reach a height of 41 meters.

The discriminate of a quadratic function is:

Δ=b²-4*a*c

In this case, being a=-5, b=2 and c=39, the discriminate is:

Δ=2²-4*(-5)*39

Δ= 4+780

Δ= 784

The discriminate is 784

The first derivative test is the process of analyzing functions using their first derivatives in order to find their extreme point.

No, ball can not reach at height of 41 meters.

Since, The height h of the ball, in meters, at time t seconds is modeled by the function  [tex]h(t)=-5t^2+2t+39[/tex]

To find maximum height where can ball reach. We use first and second derivative test.

[tex]h(t)=-5t^2+2t+39\\\\\frac{dh}{dt}=-10t+2[/tex]

Now, equate above derivative to zero.

 [tex]-10t+2=0\\\\t=1/5[/tex]

Now, find second derivative

[tex]\frac{d^{2}h }{dt^{2} } =-10[/tex]  , Which is less than zero.

Therefore, at t= 1/5 second , will give maximum height that can ball reach.

So,  h(1/5) = 39.2 meters.

That is, ball can reach at 39.2 meters.

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