How do you write a quadratic function in vertex form whose vertex \[(1,2)\] and passes through point \[(2,4)\]?
Answer
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Hint: We use the formula of vertex form of a quadratic equation and substitute the value of vertex in the required place and the point through which the quadratic equation passes at required place.
* Vertex form of a quadratic equation having vertex \[(h,k)\] and passing through the point \[(x,y)\] is given by \[y = a{(x - h)^2} + k\]
Complete step-by-step answer:
We are given the vertex for quadratic equation as \[(1,2)\]
Comparing the vertex to general vertex\[(h,k)\], we can write
$ \Rightarrow$ \[h = 1;k = 2\] … (1)
We are given the point for quadratic equation as \[(2,4)\]
Comparing the vertex to general vertex\[(x,y)\], we can write
$ \Rightarrow$\[x = 2;y = 4\] … (2)
We know that vertex form of quadratic equation having vertex \[(h,k)\]and passing through the point \[(x,y)\]is given by \[y = a{(x - h)^2} + k\]
Substitute the values of x, y, h and k from equations (1) and (2) in the formula
\[ \Rightarrow 4 = a{(2 - 1)^2} + 2\]
Subtract the values inside the bracket in right hand side of the equation
\[ \Rightarrow 4 = a{(1)^2} + 2\]
\[ \Rightarrow 4 = a + 2\]
Shift all constant values to left hand side of the equation
\[ \Rightarrow \left( {4 - 2} \right) = a\]
Subtract the values inside the bracket in right hand side of the equation
\[ \Rightarrow a = 2\]
Substitute the value of h, k and a in the equation \[y = a{(x - h)^2} + k\] and write general quadratic equation
\[ \Rightarrow y = 2{(x - 1)^2} + 2\]
\[\therefore \] The quadratic function in vertex form whose has vertex \[(1,2)\] and passes through point \[(2,4)\] is \[y = 2{(x - 1)^2} + 2\]
Note:
Many students make mistake of expanding the terms after the last step as they use the formula of \[{(a + b)^2}\] to open the right hand side and add constants which is wrong because that is not quadratic form, keep in mind using this quadratic form we can tell the vertex but if we use the expanded equation we cannot tell the vertex straight away.
* Vertex form of a quadratic equation having vertex \[(h,k)\] and passing through the point \[(x,y)\] is given by \[y = a{(x - h)^2} + k\]
Complete step-by-step answer:
We are given the vertex for quadratic equation as \[(1,2)\]
Comparing the vertex to general vertex\[(h,k)\], we can write
$ \Rightarrow$ \[h = 1;k = 2\] … (1)
We are given the point for quadratic equation as \[(2,4)\]
Comparing the vertex to general vertex\[(x,y)\], we can write
$ \Rightarrow$\[x = 2;y = 4\] … (2)
We know that vertex form of quadratic equation having vertex \[(h,k)\]and passing through the point \[(x,y)\]is given by \[y = a{(x - h)^2} + k\]
Substitute the values of x, y, h and k from equations (1) and (2) in the formula
\[ \Rightarrow 4 = a{(2 - 1)^2} + 2\]
Subtract the values inside the bracket in right hand side of the equation
\[ \Rightarrow 4 = a{(1)^2} + 2\]
\[ \Rightarrow 4 = a + 2\]
Shift all constant values to left hand side of the equation
\[ \Rightarrow \left( {4 - 2} \right) = a\]
Subtract the values inside the bracket in right hand side of the equation
\[ \Rightarrow a = 2\]
Substitute the value of h, k and a in the equation \[y = a{(x - h)^2} + k\] and write general quadratic equation
\[ \Rightarrow y = 2{(x - 1)^2} + 2\]
\[\therefore \] The quadratic function in vertex form whose has vertex \[(1,2)\] and passes through point \[(2,4)\] is \[y = 2{(x - 1)^2} + 2\]
Note:
Many students make mistake of expanding the terms after the last step as they use the formula of \[{(a + b)^2}\] to open the right hand side and add constants which is wrong because that is not quadratic form, keep in mind using this quadratic form we can tell the vertex but if we use the expanded equation we cannot tell the vertex straight away.
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