Chapter 7 - Algebra: Graphs, Functions, and Linear Systems - 7.6 Modeling Data: Exponential, Logarithmic, and Quadratic Functions - Exercise Set 7.6 - Page 477: 40

See below:

\[\left( \text{a} \right)\] In this case, the ball’s height is being modeled. When a ball is thrown upward and outward, then initially the height of the ball above the earth’s surface rises, but after some time it starts coming down. Thus, the ball’s height begins to decrease after some time. In simple language, the ball’s height will increase in the beginning and decrease thereafter. Hence, a quadratic function is a good way to handle this situation. \[\left( \text{b} \right)\] From the screen, the function can be modeled because the calculator displays the function as \[\begin{align} & y=a{{x}^{2}}+bx+c \\ & a=-0.8 \\ & b=3.2 \\ & c=6 \\ \end{align}\] Thus, the final function can be modeled by substituting the values of \[a,b,c\] in the given equation to get \[y=-0.8{{x}^{2}}+3.2x+6\] The model in function notation is \[y=-0.8{{x}^{2}}+3.2x+6\]. \[\left( \text{c} \right)\] The \[x\]-coordinate of a quadratic function \[y=a{{x}^{2}}+bx+c\] is given by \[x=-\frac{b}{2a}\]. In this case, \[a=-0.8,\ b=3.2,\ c=6\]. Hence, the \[x\]-coordinate of the quadratic function \[y=-0.8{{x}^{2}}+3.2x+6\] is given by \[\begin{align} & x=-\frac{b}{2a} \\ & =-\frac{3.2}{2\times \left( -0.8 \right)} \\ & =2 \end{align}\] Thus, the \[x\]-coordinate of the quadratic function \[y=-0.8{{x}^{2}}+3.2x+6\] is \[x=2\]. As \[a<0\], the highest peak of the quadratic function occurs at the vertex and the maximum height \[h\] is given by the value of \[y\] at \[x=2\]: \[\begin{align} & h=y\left( 2 \right) \\ & =-0.8{{\left( 2 \right)}^{2}}+3.2\left( 2 \right)+6 \\ & =-3.2+6.4+6 \\ & =9.2 \end{align}\] Hence, the maximum height of the ball occurs 2 feet from where it was thrown and the maximum height is 9.2 feet.