William Curry

2022-01-16

A ball is thrown upward and outward from a height of 6 feet. The table shows four measurements indicating the ball's height at various horizontal distances from where it was thrown. A graphing calculator displays a quadratic function that models the ball's height, y, in feet, in terms of its horizontal distance, x, in feet. Answer
$\begin{array}{|cc|}\hline x,\text{Ball's Horizontal Distance (feet)}& y,\text{Ball's Height (feet)}\\ 0& 6\\ 1& 8.1\\ 3& 6\\ 4& 2.9\\ \hline\end{array}$
a. Explain why a quadratic function was used to model the data.
In the quadratic regression screen shown in the problem statement, why is the value of the coefficient a negative?
b. Use the graphing calculator screen (shown in the box above) to express the model in function notation.
$f\left(x\right)=?$
c. Use the model from part (b) to determine the k-coordinate of the quadratic function's vertex.
The x-coordinate of the vertex is ?

Lindsey Gamble

Expert

a) According to the table above,, that height of the ball first increases and then decreases, which looks like a quadratic function.
The coefficient a in the regression screen is negative because the height of the ball increases and then decreases, hence, the quadratic must open downward.
b) Using the quadratic regression model,
$y=a{x}^{2}+bx+c$
$a=-0.9,b=2.6,c=6.1$
Hence, $y=f\left(x\right)=-0.9{x}^{2}+2.6x+6.1$
c) We have that, the vertex of the parabola $y=a{x}^{2}+bx+c$ provided by
$x=-\frac{b}{2a}$
$⇒x=-\frac{2.6}{2\left(-0.9\right)}=1.44$
The x coordinate of the vertex is 1.4
The vertex is where the height is at its highest.
Thus, ${y}_{max}=f\left(1.44\right)=-0.9{\left(1.44\right)}^{2}+2.6\left(1.44\right)+6.1=7.978$
$\approx 8$ feet
The maximum height of the ball occurs 1.4 feet from where it was thrown and the maximum height is 8 feet.

Expert

a.) After the point at which horizontal distance is 1ft, the height decreases, and decreases more rapidly with time. It appears to be a quadratic function. So, the response is A.
The graph must open downwards, as the height first increase then decrease. So, the value of the coefficient of ${x}^{2}$ must be negative. So, the response is A.
b.) $f\left(x\right)=-0.7{x}^{2}+2.1x+6.1$
c.) At vertex, the slope of the graph will be 0. Therefore, the function's differentiation will be zero. So,
$f\left(x\right)=-0.7{x}^{2}+2.1x+6.1$
${f}^{\prime }\left(x\right)=-1.4x+2.1=0$
$x=\frac{2.1}{1.4}=\frac{3}{2}=1.5$

alenahelenash

Expert

b) $y=-0.68{x}^{2}+2.05x+6.01$$y=-0.68\left({x}^{2}\frac{-205}{68}\right)$c) ${y}^{\prime }=-2\left(0.68\right)x+2.05=0$$x=\frac{2.05}{2×0.68}=1.5074$x-coordinate of vertex $=1.5074$d) $y\left(x\right)=-0.68\left(1.5074{\right)}^{2}+2.05\left(1.5074\right)+6.01$$y=7.5550$Maximum height $=7.5550$