Question

How do I find the vertex of $f(x)={{x}^{2}}+6x+5$?

Answer 1

Hint: The given equation is of the form $y=a{{x}^{2}}+bx+c$ so by comparing the equations get the values of a, b and c. Then we will find the discriminant by using the formula $D=\sqrt{{{b}^{2}}-4ac}$. Then we will use the relation $y=a{{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4a}$ where $\left( \dfrac{-b}{2a},-\dfrac{D}{4a} \right)$ is the vertex.

Complete step by step answer:
We have been given an equation $f(x)={{x}^{2}}+6x+5$.
We have to find the vertex of the given equation.
We know that if we have a quadratic equation of the form $y=a{{x}^{2}}+bx+c$ then its vertex is given as $\left( \dfrac{-b}{2a},-\dfrac{D}{4a} \right)$ where D is the discriminant which is given as $D=\sqrt{{{b}^{2}}-4ac}$.
Now, by comparing the equation with the general equation we will get
$a=1,b=6,c=5$
Now, let us find the value of discriminant. So by substituting the values in the formula we will get
$\begin{align}
  & \Rightarrow D=\sqrt{{{6}^{2}}-4\times 1\times 5} \\
 & \Rightarrow D=\sqrt{36-20} \\
 & \Rightarrow D=\sqrt{16} \\
 & \Rightarrow D=4 \\
\end{align}$
Now, substituting the values in the relation $y=a{{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4a}$ we will get
$\begin{align}
  & \Rightarrow y=1{{\left( x+\dfrac{6}{2} \right)}^{2}}-\dfrac{4}{4} \\
 & \Rightarrow y={{\left( x+3 \right)}^{2}}-1 \\
\end{align}$
Now, substituting the values in the $\left( \dfrac{-b}{2a},-\dfrac{D}{4a} \right)$ we will get
\[\Rightarrow \left( \dfrac{-6}{2\times 1},-\dfrac{4}{4\times 1} \right)\]
Now, simplifying above equation we will get
$\Rightarrow \left( -3,-1 \right)$

Hence $\left( -3,-1 \right)$ is the vertex of the given equation.

Note: We can also solve the given equation by completing the square method. For this we need to add or subtract a number from the equation such as the constant term is converted into a perfect square. Then simplify the LHS of the obtained equation to make it a square formula. Then we factor the left terms and solve the equation for x.