Q31. - Algebra 2

Question

The general form of the equation for the parabola is $y=a{x}^2+bx+c$, where $\left(x,y\right)$ is a point on the parabola. Determine the values of the $a,b,c$ for the parabola at the right. Write the general form of the equation.

Step-by-Step Solution

Answer

The general form of the parabola equation is $y=\frac{3}{2}{x}^2+3$.

Step 1 – First construct a system of equations corresponding to the given situation.

The points that lies on the parabola are $\left(0,3\right),\left(2,9\right),\left(-2,9\right)$.

 Substitute $\left(x,y\right)=\left(0,3\right)$ in $y=a{x}^2+bx+c$:

$\begin{array}{c}y=a{x}^2+bx+c\\ 3=a{\left(0\right)}^2+b\left(0\right)+c\\ 3=c\end{array}$

Substitute $\left(x,y\right)=\left(2,9\right)$ in $y=a{x}^2+bx+c$:

$\begin{array}{c}y=a{x}^2+bx+c\\ 9=a{\left(2\right)}^2+b\left(2\right)+c\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute 2 for }x,9\text{ for }y\\ 9=4a+2b+c\end{array}$

Substitute $\left(x,y\right)=\left(-2,9\right)$ in $y=a{x}^2+bx+c$:

$\begin{array}{c}y=a{x}^2+bx+c\\ 9=a{(-2)}^2+b(-2)+c\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute }-\text{2 for }x,9\text{ for }y\\ 9=4a-2b+c\end{array}$

So, the system of equations will be:

$\begin{array}{d}4a-2b+c=9\\ 4a+2b+c=9\\ c=3\end{array}$

Step 2– Use the elimination method to solve the system of equations in two variables.

Substitute $c=3$ in $4a-2b+c=9$:

$\begin{array}{c}4a-2b+c=9\\ 4a-2b+3=9\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute 3 for }c\\ 4a-2b=6\end{array}$

Substitute $c=3$ in $4a+2b+c=9$

$\begin{array}{c}4a+2b+c=9\\ 4a+2b+3=9\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute 3 for }c\\ 4a+2b=6\end{array}$

:Add the equation$4a-2b=6$ to the equation $4a+2b=6$

$\begin{array}{l}\underset{\_}{\begin{array}{l}4a-2b=6\\ 4a+2b=6\end{array}}\\ 8a=12\end{array}$

.So, the resultant equation is $8a=12$.

Step 3 – Find the values of a and b .

Solve $8a=12$ for $a$:

$\begin{array}{c}8a=12\\ \frac{8a}{8}=\frac{12}{8}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}divide both sides by }8\\ a=\frac{3}{2}\end{array}$

Substitute $a=\frac{3}{2}$ in $4a+2b=6$  and find the value of $b$.

$\begin{array}{c}4a+2b=6\\ 4\left(\frac{3}{2}\right)+2b=6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute }\frac{3}{2}\text{ for }a\\ 6+2b=6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Simplify}\\ 2b=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract }6\text{ from both sides}\\ b=0\end{array}$

Hence, the solution of the given system of equations is$\left(a,b,c\right)=\left(\frac{3}{2},0,3\right)$.

So, the general form of the parabola equation is $y=\frac{3}{2}{x}^2+3$.