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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+t+4

will the ball reach a height of 5 meters?

Respuesta :

Step-by-step explanation:

Proving by Vertex Form:Yes, if we put this equation in vertex form

[tex]a(x - h) {}^{2} + k[/tex]

where h,k is vertex

we can find if it reaches 5.

First we do -b/2a.

[tex] \frac{ - 1}{10} [/tex]

Is a solution, if we plug that in we get

Which gives

[tex]3.95[/tex]

So the vertex form is

[tex]5(x + 0.1) {}^{2} + 3.95[/tex]

Since the graph is facing upwards,because a is positive, the function will pass all values greater than 3.95, and 5 is one of them

Second Way: Watch if you want to see how long is reaches 5 meters.

Set the quadratic equation equal to 5.

[tex]5 {t}^{2} + t + 4 = 5[/tex]

[tex] {5t}^{2} + t - 1 = 0[/tex]

[tex]5 {t}^{2} + t - 1 = 0[/tex]

Do quadratic formula, which is

[tex] - b± \frac{ \sqrt{b {}^{2} - 4ac} }{2a} [/tex]

So a is 5 b is 1 and c is -1 so we get

[tex] - 1± \frac{ \sqrt{1 {}^{2} - 4( 5)( - 1)} }{2(5)} [/tex]

[tex] - 1± \frac{ \sqrt{1 + 20} }{10} [/tex]

[tex] - 1± \frac{ \sqrt{21} }{10} [/tex]

[tex] \frac{ - 1}{10} + \frac{ \sqrt{21} }{10} [/tex]

and

[tex] \frac{ - 1}{10} - \frac{ \sqrt{21} }{10} [/tex]

are the solutions but since time is scalar, it can't be negative so qe take the positive solution

[tex] \frac{ - 1}{10} + \frac{ \sqrt{21} }{10} [/tex]

which is approximate 0.358.

So if you wanted to know, it will take 0.36 seconds to reach a height of 5 meters.