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

How do I find the vertex of f(x)=x2+6x+5f(x)={{x}^{2}}+6x+5?

Explanation

Solution

The given equation is of the form y=ax2+bx+cy=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=b2−4acD=\sqrt{{{b}^{2}}-4ac}. Then we will use the relation y=a(x+b2a)2−D4ay=a{{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4a} where (−b2a,−D4a)\left( \dfrac{-b}{2a},-\dfrac{D}{4a} \right) is the vertex.

Complete step by step answer:
We have been given an equation f(x)=x2+6x+5f(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=ax2+bx+cy=a{{x}^{2}}+bx+c then its vertex is given as (−b2a,−D4a)\left( \dfrac{-b}{2a},-\dfrac{D}{4a} \right) where D is the discriminant which is given as D=b2−4acD=\sqrt{{{b}^{2}}-4ac}.
Now, by comparing the equation with the general equation we will get
a=1,b=6,c=5a=1,b=6,c=5
Now, let us find the value of discriminant. So by substituting the values in the formula we will get
⇒D=62−4×1×5 ⇒D=36−20 ⇒D=16 ⇒D=4 \begin{aligned} & \Rightarrow D=\sqrt{{{6}^{2}}-4\times 1\times 5} \\\ & \Rightarrow D=\sqrt{36-20} \\\ & \Rightarrow D=\sqrt{16} \\\ & \Rightarrow D=4 \\\ \end{aligned}
Now, substituting the values in the relation y=a(x+b2a)2−D4ay=a{{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4a} we will get
⇒y=1(x+62)2−44 ⇒y=(x+3)2−1 \begin{aligned} & \Rightarrow y=1{{\left( x+\dfrac{6}{2} \right)}^{2}}-\dfrac{4}{4} \\\ & \Rightarrow y={{\left( x+3 \right)}^{2}}-1 \\\ \end{aligned}
Now, substituting the values in the (−b2a,−D4a)\left( \dfrac{-b}{2a},-\dfrac{D}{4a} \right) we will get
⇒(−62×1,−44×1)\Rightarrow \left( \dfrac{-6}{2\times 1},-\dfrac{4}{4\times 1} \right)
Now, simplifying above equation we will get
⇒(−3,−1)\Rightarrow \left( -3,-1 \right)

Hence (−3,−1)\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.