
How do you rewrite \[y={{x}^{2}}+14x+29\] in vertex form ?
Answer
547.2k+ views
Hint:In the given question, we have been asked to write a given equation in the vertex form. In order to write the equation in a vertex form, we need to use the vertex form of the equation; i.e.\[y=a{{\left( x-h \right)}^{2}}-k\] where, \[a\] equals to the coefficient of \[{{x}^{2}}\], ‘h’ is the x-coordinate of the vertex, ‘k’ is the y-coordinate of the vertex. First we need to solve for the coordinate of the vertex and then putting all the values we will get our equation in vertex form.
Formula used:
The vertex form of the equation; i.e.
\[y=a{{\left( x-h \right)}^{2}}-k\]
Where, \[a\] equals to the coefficient of \[{{x}^{2}}\], ‘h’ is the x-coordinate of the vertex and ‘k’ is the y-coordinate of the vertex.
Complete step by step answer:
We have given that,
\[\Rightarrow y={{x}^{2}}+14x+29\]
Using the vertex form of the equation; i.e.
\[y=a{{\left( x-h \right)}^{2}}-k\]
Where, \[a\] equals to the coefficient of \[{{x}^{2}}\] and ‘h’ is the x-coordinate of the vertex.‘k’ is the y-coordinate of the vertex.
Vertex of the given equation \[y={{x}^{2}}+14x+29\] is;
Now, solving for the value of ‘x’ and ‘y’,
\[\Rightarrow y={{x}^{2}}+14x+29\]
Here ‘a’ is the coefficient of first term and b is the coefficient of ‘x’.
\[\Rightarrow x=dfrac{-b}{2a}=dfrac{-14}{2}=-7\]
\[\Rightarrow x=-7\]
Now, solving for the value of ‘y’.
Putting x = -7 in the given equation, we get
\[\Rightarrow y={{\left( -7 \right)}^{2}}+14\left( -7 \right)+29\]
Simplifying the above equation, we get
\[\Rightarrow y=49-98+29=78-98=-20\]
\[\Rightarrow y=-20\]
Therefore, the vertex\[(x,y)=(-7,-20)\].
Thus,
Vertex = (-7, -20), ‘a’ = 1
Now, substituting these values in the vertex equation form, we obtain
\[y=a{{\left( x-h \right)}^{2}}-k\]
\[\Rightarrow y=1{{\left( x-\left( -7 \right) \right)}^{2}}-\left( -20 \right)\]
Simplifying the above, we get
\[\therefore y={{\left( x+7 \right)}^{2}}+20\]
Hence, the equation \[y={{x}^{2}}+14x+29\] in vertex form is \[y={{\left( x+7 \right)}^{2}}+20\].
Note:Students should be very well aware of the vertex form of the equation. Also, they should know about the concept of converting the given equation into vertex form. They should also be very careful while doing the calculations to avoid making errors. Students should also know to find the coordinates of the vertex.
Formula used:
The vertex form of the equation; i.e.
\[y=a{{\left( x-h \right)}^{2}}-k\]
Where, \[a\] equals to the coefficient of \[{{x}^{2}}\], ‘h’ is the x-coordinate of the vertex and ‘k’ is the y-coordinate of the vertex.
Complete step by step answer:
We have given that,
\[\Rightarrow y={{x}^{2}}+14x+29\]
Using the vertex form of the equation; i.e.
\[y=a{{\left( x-h \right)}^{2}}-k\]
Where, \[a\] equals to the coefficient of \[{{x}^{2}}\] and ‘h’ is the x-coordinate of the vertex.‘k’ is the y-coordinate of the vertex.
Vertex of the given equation \[y={{x}^{2}}+14x+29\] is;
Now, solving for the value of ‘x’ and ‘y’,
\[\Rightarrow y={{x}^{2}}+14x+29\]
Here ‘a’ is the coefficient of first term and b is the coefficient of ‘x’.
\[\Rightarrow x=dfrac{-b}{2a}=dfrac{-14}{2}=-7\]
\[\Rightarrow x=-7\]
Now, solving for the value of ‘y’.
Putting x = -7 in the given equation, we get
\[\Rightarrow y={{\left( -7 \right)}^{2}}+14\left( -7 \right)+29\]
Simplifying the above equation, we get
\[\Rightarrow y=49-98+29=78-98=-20\]
\[\Rightarrow y=-20\]
Therefore, the vertex\[(x,y)=(-7,-20)\].
Thus,
Vertex = (-7, -20), ‘a’ = 1
Now, substituting these values in the vertex equation form, we obtain
\[y=a{{\left( x-h \right)}^{2}}-k\]
\[\Rightarrow y=1{{\left( x-\left( -7 \right) \right)}^{2}}-\left( -20 \right)\]
Simplifying the above, we get
\[\therefore y={{\left( x+7 \right)}^{2}}+20\]
Hence, the equation \[y={{x}^{2}}+14x+29\] in vertex form is \[y={{\left( x+7 \right)}^{2}}+20\].
Note:Students should be very well aware of the vertex form of the equation. Also, they should know about the concept of converting the given equation into vertex form. They should also be very careful while doing the calculations to avoid making errors. Students should also know to find the coordinates of the vertex.
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