Question

What is the vertex form of \[y=2{{x}^{2}}+8x-3\]?

Answer 1

Hint: For solving this question you should know about the vertex form of quadratic equations. By the concept of vertex form representation of quadratic equations it is clear that we can easily calculate the vertex form of any function (quadratic) by its standard equation. And if we see that the standard form of quadratic equation is \[y=a{{x}^{2}}+bx+c\] and the vertex form for the quadratic equations is \[y=a{{\left( x-h \right)}^{2}}+k\].

Complete step by step answer:
According to our question it is asked to us to calculate the vertex form of \[y=2{{x}^{2}}+8x-3\].
The equation of a parabola in vertex form is \[y=a{{\left( x-h \right)}^{2}}+k\], where h, k are the coordinates of the vertex and a is a multiplier. To obtain the vertex form use completing the square.
The co – efficient of the \[{{x}^{2}}\] term must be 1.
And add / subtract (\[\dfrac{1}{2}\] coefficient of x term) \[^{2}\] to the common term of \[{{x}^{2}}+bx\].
We can understand it by an example:
Eg., \[y=3{{x}^{2}}+9x-4\]
Step 1: First we have to make the coefficient of the \[{{x}^{2}}\] term must be 1.
So, \[y=3\left( {{x}^{2}}+3x \right)-4\]
Now, Step 2: add/ subtract (\[\dfrac{1}{2}\] coefficient of x term) \[^{2}\] to \[{{x}^{2}}+3x\]
\[\begin{align}
  & y=3\left( {{x}^{2}}+2.\left( \dfrac{3}{2} \right)x+\dfrac{9}{4}-\dfrac{9}{4} \right)-4 \\
 & y=3{{\left( x+\dfrac{3}{2} \right)}^{2}}-\dfrac{27}{4}-4 \\
 & y=3{{\left( x+\dfrac{3}{2} \right)}^{2}}-\dfrac{43}{4} \\
\end{align}\]
So, it is it’s vertex form.
If we see our question then, \[y=2{{x}^{2}}+8x-3\]
To make the coefficient of \[{{x}^{2}}\] must be 1.
\[\Rightarrow y=2\left( {{x}^{2}}+4x \right)-3\]
Now add/ subtract (\[\dfrac{1}{2}\] coefficient of x term) \[^{2}\] to \[{{x}^{2}}+4x\]
\[\begin{align}
  & \Rightarrow y=2\left( {{x}^{2}}+2\left( 2 \right)x+4-4 \right)-3 \\
 & \Rightarrow y=2{{\left( x+2 \right)}^{2}}-8-3 \\
 & \Rightarrow y=2{{\left( x+2 \right)}^{2}}-11 \\
\end{align}\]

So, the vertex form of \[y=2{{x}^{2}}+8x-3\] is \[y=2{{\left( x+2 \right)}^{2}}-11\].

Note: During calculating the vertex form of any quadratic equation we are always careful for the coefficient of \[{{x}^{2}}\]. It will always remain 1. If any other is given then remove it by taking common. And always make the whole square of this term because if it will not become whole square then it can’t change in the vertex form.