Answer

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Hint: The general form of the parabola \[y=a{{x}^{2}}+bx+c\] is to be used while solving this question.

Complete step-by-step answer:

In the question, it is given that the axis of the parabola is parallel to the $y-$axis. So, the parabola would look like the figure below.

We know that the general formula for this form of parabola is given by,

\[y=a{{x}^{2}}+bx+c\ldots \ldots \ldots \left( i \right)\]

It is given in the question that the parabola passes through the points \[\left( 0,2 \right),\left( -1,0 \right)\] and \[\left( 1,6 \right)\]. Since the parabola passes through the three points, we know that these points must satisfy the equation of the parabola. So, we can substitute each point in the equation $\left( i \right)$ and formulate three sets of equations.

Considering the first point \[\left( 0,2 \right)\] and substituting the values of \[x=0,y=2\] in equation $\left( i \right)$, we get

\[\begin{align}

& 2=a\times 0+b\times 0+c \\

& c=2\ldots \ldots \ldots \left( ii \right) \\

\end{align}\]

Considering the second point \[\left( -1,0 \right)\] and substituting the values of \[x=-1,y=0\] in equation $\left( i \right)$, we get

\[\begin{align}

& 0=a\times {{\left( -1 \right)}^{2}}+b\times -1+c \\

& 0=a-b+c \\

\end{align}\]

Now substituting the value of \[c\] from equation $\left( ii \right)$ in the above equation, we get

\[\begin{align}

& 0=a-b+2 \\

& b-a=2\ldots \ldots \ldots \left( iii \right) \\

\end{align}\]

Considering the last point \[\left( 1,6 \right)\] and substituting the values of \[x=1,y=6\] in equation $\left( i \right)$, we get

$\begin{align}

& 6=a\times {{1}^{2}}+b\times 1+c \\

& 6=a+b+c \\

\end{align}$

Now substituting the value of \[c\] from equation $\left( ii \right)$ in the above equation, we get

\[\begin{align}

& 6=a+b+2 \\

& a+b=4\ldots \ldots \ldots \left( iv \right) \\

\end{align}\]

We have two equations \[\left( iii \right)\] and $\left( iv \right)$ to get the values of the \[a\] and $b$. So, adding the equations,

\[\dfrac{\begin{align}

& b-a=2 \\

& a+b=4 \\

\end{align}}{\begin{align}

& 2b=6 \\

& b=3 \\

\end{align}}\]

Substituting \[b=3\] in equation \[\left( iii \right)\], we get

\[\begin{align}

& 3-a=2 \\

& a=1 \\

\end{align}\]

Now we have the values as $a=1,b=3,c=2$. So, we can substitute this in equation $\left( i \right)$,

\[y={{x}^{2}}+3x+2\]

Therefore, the required equation of the parabola is obtained as \[y={{x}^{2}}+3x+2\].

Note: As three points are given in the question, we can formulate three equations and easily compute the three unknowns in the equation. If you are familiar with the cross-multiplication method, you can solve the equations and get the values of \[a,b,c\] in less time.

Complete step-by-step answer:

In the question, it is given that the axis of the parabola is parallel to the $y-$axis. So, the parabola would look like the figure below.

We know that the general formula for this form of parabola is given by,

\[y=a{{x}^{2}}+bx+c\ldots \ldots \ldots \left( i \right)\]

It is given in the question that the parabola passes through the points \[\left( 0,2 \right),\left( -1,0 \right)\] and \[\left( 1,6 \right)\]. Since the parabola passes through the three points, we know that these points must satisfy the equation of the parabola. So, we can substitute each point in the equation $\left( i \right)$ and formulate three sets of equations.

Considering the first point \[\left( 0,2 \right)\] and substituting the values of \[x=0,y=2\] in equation $\left( i \right)$, we get

\[\begin{align}

& 2=a\times 0+b\times 0+c \\

& c=2\ldots \ldots \ldots \left( ii \right) \\

\end{align}\]

Considering the second point \[\left( -1,0 \right)\] and substituting the values of \[x=-1,y=0\] in equation $\left( i \right)$, we get

\[\begin{align}

& 0=a\times {{\left( -1 \right)}^{2}}+b\times -1+c \\

& 0=a-b+c \\

\end{align}\]

Now substituting the value of \[c\] from equation $\left( ii \right)$ in the above equation, we get

\[\begin{align}

& 0=a-b+2 \\

& b-a=2\ldots \ldots \ldots \left( iii \right) \\

\end{align}\]

Considering the last point \[\left( 1,6 \right)\] and substituting the values of \[x=1,y=6\] in equation $\left( i \right)$, we get

$\begin{align}

& 6=a\times {{1}^{2}}+b\times 1+c \\

& 6=a+b+c \\

\end{align}$

Now substituting the value of \[c\] from equation $\left( ii \right)$ in the above equation, we get

\[\begin{align}

& 6=a+b+2 \\

& a+b=4\ldots \ldots \ldots \left( iv \right) \\

\end{align}\]

We have two equations \[\left( iii \right)\] and $\left( iv \right)$ to get the values of the \[a\] and $b$. So, adding the equations,

\[\dfrac{\begin{align}

& b-a=2 \\

& a+b=4 \\

\end{align}}{\begin{align}

& 2b=6 \\

& b=3 \\

\end{align}}\]

Substituting \[b=3\] in equation \[\left( iii \right)\], we get

\[\begin{align}

& 3-a=2 \\

& a=1 \\

\end{align}\]

Now we have the values as $a=1,b=3,c=2$. So, we can substitute this in equation $\left( i \right)$,

\[y={{x}^{2}}+3x+2\]

Therefore, the required equation of the parabola is obtained as \[y={{x}^{2}}+3x+2\].

Note: As three points are given in the question, we can formulate three equations and easily compute the three unknowns in the equation. If you are familiar with the cross-multiplication method, you can solve the equations and get the values of \[a,b,c\] in less time.

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