Answer:
The standard form of a quadratic equation is given below as
[tex]y=ax^2+bx+c[/tex]The y-intercept given is
[tex](0,-1.4)[/tex]This indicates that at x=0, y= 1.4
Substitute x=0, y= 1.4 in the quadratic expression below
[tex]\begin{gathered} y=ax^2+bx+c \\ a(0)^2+b(0)+c=-1.4 \\ c=1.4 \end{gathered}[/tex]The x-intercept is given below as
[tex](0.905,0)[/tex]This means at x=0.905,y=0
Substitute x=0.905,y=0 in the quadratic expression above
[tex]\begin{gathered} y=ax^2+bx+c \\ 0=a(0.905)^2+b(0.905)+c \\ \text{recall that:} \\ c=-1.4 \\ 0=a(0.905)^2+b(0.905)+c \\ 0=a(0.905)^2+b(0.905)-1.4 \\ 0.819a+0.905b=1.4----(1) \end{gathered}[/tex]Since the third points below pass through the parabola, the third point is given below as
[tex](3.07,0.314)[/tex]Substitute when x=3.07,y=0.314 in the quadratic expression above
[tex]\begin{gathered} y=ax^2+bx+c \\ a(3.07)^2+b(3.07)-1.4=0.314 \\ 9.425a+3.07b=0.314+1.4 \\ 9.425a+3.07b=1.714-----(2) \end{gathered}[/tex]Combine equations (1) and (2) and solve simultaneously
[tex]\begin{gathered} 0.819a+0.905b=1.4 \\ 9.425a+3.07b=1.714 \\ \text{mulipy equation (1) by 9.425 and equation 2 by 0.819} \\ 7.719a+8.53b=13.195 \\ 7.719a+2.514b=1.404 \\ \text{substracting both equations, we will have} \end{gathered}[/tex][tex]\begin{gathered} 7.719a-7.719a+8.53b-2.514b=13.195-1.404 \\ \frac{6.016b}{6.016}=\frac{11.791}{6.016} \\ b=1.96 \end{gathered}[/tex]Substitute the value of b=1.96 in the equation (1)
[tex]\begin{gathered} 0.819a+0.905b=1.4 \\ 0.819a+0.905(1.96)=1.4 \\ 0.819a+1.7738=1.4 \\ 0.819a=1.4-1.7738 \\ \frac{0.819a}{0.819}=\frac{-0.3738}{0.819} \\ a=-0.46 \end{gathered}[/tex]Hence,
The general equation will be
[tex]y=-0.46x^2+1.96x-1.4[/tex]
