
How do you write the quadratic equation function \[y = {x^2} - 4x + 15\] in vertex form?
Answer
533.1k+ views
Hint: Here, we will first take a value common from all the terms of the equation. Then we will use the completing the square method to simplify the equation. Then we will use the general equation of vertex form to write the obtained equation in vertex form.
Complete step-by-step answer:
The quadratic equation given to us is \[y = {x^2} - 4x + 15\].
Now, taking 1 common in right hand side from the first two values we get,
\[ \Rightarrow y = 1\left( {{x^2} - 4x} \right) + 15\]
Next, we will complete the square of the bracket term having variable \[x\].
So, we will add and subtract 4 in the bracket having variable \[x\] to get.
\[ \Rightarrow y = 1\left( {{x^2} - 4x + 4 - 4} \right) + 15\]
Simplifying the equation further, we get
\[\begin{array}{l} \Rightarrow y = 1\left( {{x^2} - 4x + 4} \right) - 4 \times 1 + 15\\ \Rightarrow y = 1\left( {{x^2} - 4x + 4} \right) - 4 + 15\end{array}\]
Subtracting the like terms, we get
\[ \Rightarrow y = 1\left( {{x^2} - 4x + 4} \right) + 11\]
Now using the identity \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\], we get
\[ \Rightarrow y = 1{\left( {x - 2} \right)^2} + 11\]…..\[\left( 1 \right)\]
We know that the vertex form of a quadratic equation is \[y = m{\left( {x - a} \right)^2} + b\].
As equation \[\left( 1 \right)\] is the vertex form, we will not make any changes in it.
Hence vertex form of \[y = {x^2} - 4x + 15\] is \[y = 1{\left( {x - 2} \right)^2} + 11\] where vertex is \[\left( {2,11} \right)\].
Note:
The vertex form of a quadratic equation is \[y = m{\left( {x - a} \right)^2} + b\] where the slope of the equation is \[m\] and \[a,b\] is the vertex of \[x,y\] respectively. The vertex form is an alternative way to write the equation of a parabola. We need to keep in mind that in vertex form the value of \[a\] is subtracted and the value of \[b\] is added. The common mistake we can make in vertex form is of the sign one should always double check the positive and negative sign in this type of form and don’t get confused in what is the value of the vertex. Vertex form is very useful when we are dealing with parabola figure questions.
Complete step-by-step answer:
The quadratic equation given to us is \[y = {x^2} - 4x + 15\].
Now, taking 1 common in right hand side from the first two values we get,
\[ \Rightarrow y = 1\left( {{x^2} - 4x} \right) + 15\]
Next, we will complete the square of the bracket term having variable \[x\].
So, we will add and subtract 4 in the bracket having variable \[x\] to get.
\[ \Rightarrow y = 1\left( {{x^2} - 4x + 4 - 4} \right) + 15\]
Simplifying the equation further, we get
\[\begin{array}{l} \Rightarrow y = 1\left( {{x^2} - 4x + 4} \right) - 4 \times 1 + 15\\ \Rightarrow y = 1\left( {{x^2} - 4x + 4} \right) - 4 + 15\end{array}\]
Subtracting the like terms, we get
\[ \Rightarrow y = 1\left( {{x^2} - 4x + 4} \right) + 11\]
Now using the identity \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\], we get
\[ \Rightarrow y = 1{\left( {x - 2} \right)^2} + 11\]…..\[\left( 1 \right)\]
We know that the vertex form of a quadratic equation is \[y = m{\left( {x - a} \right)^2} + b\].
As equation \[\left( 1 \right)\] is the vertex form, we will not make any changes in it.
Hence vertex form of \[y = {x^2} - 4x + 15\] is \[y = 1{\left( {x - 2} \right)^2} + 11\] where vertex is \[\left( {2,11} \right)\].
Note:
The vertex form of a quadratic equation is \[y = m{\left( {x - a} \right)^2} + b\] where the slope of the equation is \[m\] and \[a,b\] is the vertex of \[x,y\] respectively. The vertex form is an alternative way to write the equation of a parabola. We need to keep in mind that in vertex form the value of \[a\] is subtracted and the value of \[b\] is added. The common mistake we can make in vertex form is of the sign one should always double check the positive and negative sign in this type of form and don’t get confused in what is the value of the vertex. Vertex form is very useful when we are dealing with parabola figure questions.
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