Question

an arrow is fired into the air with an initial velocity of 64 feet per second. the height in feet of the arrow t seconds after it was shot into the air is given by the function ( h(t)=-16t^{2}+64t ). find the maximum height of the arrow.
96 ft
64 ft
32 ft
192 ft
question 3
solve the system by the substitution method.
( x + y = 32 )
( y = x^{2}-4x + 4 )
( { (2,30)} )
( { (-7,39),(4,28)} )
( { (7,25),(-4,36)} )
( { (7,39),(-4,36)} )

Explanation:

Step1: Identify the vertex of the quadratic function

For a quadratic function \(y = ax^{2}+bx + c\), the \(x\) - coordinate of the vertex is given by \(t=-\frac{b}{2a}\). In the function \(h(t)=-16t^{2}+64t\), \(a=-16\) and \(b = 64\).

$$t=-\frac{64}{2\times(-16)}$$

$$t = 2$$

Step2: Find the maximum height

Substitute \(t = 2\) into the function \(h(t)\).

$$h(2)=-16\times(2)^{2}+64\times2$$

$$h(2)=-16\times4 + 128$$

$$h(2)=-64+128$$

$$h(2)=64$$

for the system of equations:

Step1: Substitute \(y\) in the first equation

Substitute \(y=x^{2}-4x + 4\) into \(x + y=32\).

$$x+(x^{2}-4x + 4)=32$$

$$x^{2}-3x+4 - 32=0$$

$$x^{2}-3x-28=0$$

Step2: Factor the quadratic equation

Factor \(x^{2}-3x - 28=(x - 7)(x+4)=0\)

Step3: Solve for \(x\)

Set each factor equal to zero: \(x-7 = 0\) gives \(x = 7\); \(x + 4=0\) gives \(x=-4\)

Step4: Solve for \(y\)

When \(x = 7\), \(y=32 - 7=25\)
When \(x=-4\), \(y=32-(-4)=36\)

Answer:

64 ft