A ball is thrown upward and outward from a height of 6 feet. The table shows four measurements indic

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, 
${\displaystyle y=a{x}^2+bx+c}$ 
${\displaystyle a=-0.9,b=2.6,c=6.1}$ 
Hence, ${\displaystyle y=f\left(x\right)=-0.9x^2+2.6x+6.1}$ 
c) We have that, the vertex of the parabola ${\displaystyle y=a{x}^2+bx+c}$ provided by
${\displaystyle x=-\frac{b}{2a}}$ 
${\displaystyle \Rightarrow x=-\frac{2.6}{2(-0.9)}=1.44}$ 
The x coordinate of the vertex is 1.4 
The vertex is where the height is at its highest.
Thus, ${\displaystyle y_{max}=f\left(1.44\right)=-0.9{\left(1.44\right)}^2+2.6\left(1.44\right)+6.1=7.978}$ 
${\displaystyle \approx 8}$ feet 
The maximum height of the ball occurs 1.4 feet from where it was thrown and the maximum height is 8 feet.

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 ${\displaystyle x^2}$ must be negative. So, the response is A.
b.) ${\displaystyle f\left(x\right)=-0.7x^2+2.1x+6.1}$
c.) At vertex, the slope of the graph will be 0. Therefore, the function's differentiation will be zero. So,
${\displaystyle f\left(x\right)=-0.7x^2+2.1x+6.1}$
${\displaystyle f^{\prime }\left(x\right)=-1.4x+2.1=0}$
${\displaystyle x=\frac{2.1}{1.4}=\frac{3}{2}=1.5}$