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The height, 
h, in feet, of a baseball 
t seconds after Tobin hit it with a baseball bat can be modeled by the equation:

h(t)=-16t^(2)+64 t+4
Which of the following equivalent expressions displays the value of the baseball's maximum height as a constant or coefficient?
Choose 1 answer:
(A) 
-16(t-2)^(2)+68
(B) 
-4t(4t-16)+4
(C) 
-16(t^(2)-4t-(1)/(4))
(D) 
-16(t^(˙)^(2)-4t)+4

The height, h h , in feet, of a baseball t t seconds after Tobin hit it with a baseball bat can be modeled by the equation:\newlineh(t)=16t2+64t+4 h(t)=-16 t^{2}+64 t+4 \newlineWhich of the following equivalent expressions displays the value of the baseball's maximum height as a constant or coefficient?\newlineChoose 11 answer:\newline(A) 16(t2)2+68 -16(t-2)^{2}+68 \newline(B) 4t(4t16)+4 -4 t(4 t-16)+4 \newline(C) 16(t24t14) -16\left(t^{2}-4 t-\frac{1}{4}\right) \newline(D) 16(t24t)+4 -16\left(t^{2}-4 t\right)+4

Full solution

Q. The height, h h , in feet, of a baseball t t seconds after Tobin hit it with a baseball bat can be modeled by the equation:\newlineh(t)=16t2+64t+4 h(t)=-16 t^{2}+64 t+4 \newlineWhich of the following equivalent expressions displays the value of the baseball's maximum height as a constant or coefficient?\newlineChoose 11 answer:\newline(A) 16(t2)2+68 -16(t-2)^{2}+68 \newline(B) 4t(4t16)+4 -4 t(4 t-16)+4 \newline(C) 16(t24t14) -16\left(t^{2}-4 t-\frac{1}{4}\right) \newline(D) 16(t24t)+4 -16\left(t^{2}-4 t\right)+4
  1. Problem Understanding: Understand the problem.\newlineWe are given the quadratic equation h(t)=16t2+64t+4h(t) = -16t^2 + 64t + 4, which models the height of a baseball tt seconds after being hit. We need to find an equivalent expression that shows the maximum height of the baseball as a constant or coefficient.
  2. Quadratic Equation Structure: Recognize the structure of the quadratic equation.\newlineThe given equation is in the standard form of a quadratic equation, which is h(t)=at2+bt+ch(t) = at^2 + bt + c. The maximum height of the baseball will occur at the vertex of the parabola represented by this quadratic equation.
  3. Finding Time of Maximum Height: Find the time at which the maximum height occurs.\newlineThe time at which the maximum height occurs can be found using the formula t=b2at = -\frac{b}{2a}, where aa is the coefficient of t2t^2 and bb is the coefficient of tt in the quadratic equation.\newlineFor our equation, a=16a = -16 and b=64b = 64, so t=642×16=2t = -\frac{64}{2 \times -16} = 2 seconds.
  4. Writing Equation in Vertex Form: Write the equation in vertex form.\newlineThe vertex form of a quadratic equation is h(t)=a(th)2+kh(t) = a(t - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola. Since we know the time of the maximum height is 22 seconds, we can write the equation as h(t)=16(t2)2+kh(t) = -16(t - 2)^2 + k.
  5. Calculating Maximum Height: Calculate the maximum height kk. To find the value of kk, we substitute t=2t = 2 into the original equation h(t)=16t2+64t+4h(t) = -16t^2 + 64t + 4. h(2)=16(2)2+64(2)+4=64+128+4=68h(2) = -16(2)^2 + 64(2) + 4 = -64 + 128 + 4 = 68 feet. So, the maximum height kk is 6868 feet.
  6. Final Vertex Form Equation: Write the final vertex form equation.\newlineNow that we have the value of kk, we can write the final vertex form equation as h(t)=16(t2)2+68h(t) = -16(t - 2)^2 + 68.
  7. Matching with Given Choices: Match the final vertex form equation with the given choices.\newlineThe final vertex form equation we found is h(t)=16(t2)2+68h(t) = -16(t - 2)^2 + 68, which matches choice (A).

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