Question

How do you find the vertex of the parabola$y=3{{x}^{2}}+5x+8$?

Answer 1

Hint:The above given equation $y=3{{x}^{2}}+5x+8$ is a parabolic equation. A parabola is a U-shaped plane curve where any point is at an equal distance from a fixed point which is also known as focus and from a fixed straight line which is known as the directrix. The standard parabolic equation is always represented as $y=a{{x}^{2}}+bx+c$.

Complete step by step solution:
The given equation is:
$\Rightarrow y=3{{x}^{2}}+5x+8$
Here we have to find the vertex of the parabola. The standard parabolic equation is $y=a{{x}^{2}}+bx+c$ and to find the value of $y$ first we have to find the x-coordinate of the equation. We know the formula to find the x-coordinate which is $\dfrac{-b}{2a}$ . And we substitute x-coordinate in the given equation we get the value of y-coordinate.
Now compare the given equation with the standard parabolic equation we get,
$\Rightarrow a=3,b=5,c=8$
Now put the above values in the formula of x-coordinate, we get
$\Rightarrow x=\dfrac{-5}{2\left( 3 \right)}$
By more simplifying we get,
$\Rightarrow x=\dfrac{-5}{6}$
Now we will put this value of x in the given equation we get,
$\begin{align}
  & \Rightarrow y=3{{\left( \dfrac{-5}{6} \right)}^{2}}+5\left( \dfrac{-5}{6} \right)+8 \\
 & \Rightarrow y=3\left( \dfrac{25}{36} \right)-\dfrac{25}{6}+8 \\
\end{align}$
More solving it then we get
$\begin{align}
  & \Rightarrow y=\dfrac{25}{12}-\dfrac{25}{6}+8 \\
 & \Rightarrow y=\dfrac{25-50+96}{12} \\
 & \Rightarrow y=\dfrac{71}{12} \\
\end{align}$
Hence we get the vertex of the parabola which is $\left( \dfrac{-5}{6},\dfrac{71}{12} \right)$.

Note: we can also solve the above parabolic equation by using another way.
The given parabola equation is:
$\Rightarrow y=3{{x}^{2}}+5x+8$
Now we will simplify the above equation as:
$\Rightarrow y=3\left( {{x}^{2}}+\dfrac{5}{3}x \right)+8$
Now add and subtract ${{\left( \dfrac{5}{6} \right)}^{2}}$ in the above equation we get,
$\begin{align}
  & \Rightarrow y=3\left( {{x}^{2}}+2\left( \dfrac{5}{6} \right)x+{{\left( \dfrac{5}{6} \right)}^{2}} \right)-3{{\left( \dfrac{5}{6} \right)}^{2}}+8 \\
 & \Rightarrow y=3{{\left( x+\dfrac{5}{6} \right)}^{2}}+\dfrac{71}{12} \\
\end{align}$
Now add and subtract $\dfrac{71}{12}$ in both side of the equation, we get,
$\begin{align}
  & \Rightarrow y-\dfrac{71}{12}=3{{\left( x+\dfrac{5}{6} \right)}^{2}} \\
 & \Rightarrow \dfrac{1}{3}\left( y-\dfrac{71}{12} \right)={{\left( x+\dfrac{5}{6} \right)}^{2}} \\
\end{align}$
Now we know the above equation is looking like an upward parabola equation:
$\Rightarrow \left( x-{{x}_{1}} \right)=4a\left( y-{{y}_{1}} \right)$
And which has vertex as:
$\Rightarrow \left( x-{{x}_{1}} \right)=0,\left( y-{{y}_{1}} \right)=0$
By comparing the above equation with$\dfrac{1}{3}\left( y-\dfrac{71}{12} \right)={{\left( x+\dfrac{5}{6} \right)}^{2}}$, we get
$\begin{align}
  & \Rightarrow \left( x+\dfrac{5}{6}=0,y-\dfrac{71}{12}=0 \right) \\
 & \Rightarrow \left( x=\dfrac{-5}{6},y=\dfrac{71}{12} \right) \\
\end{align}$
Hence we get the same vertex as we get above$\left( \dfrac{-5}{6},\dfrac{71}{12} \right)$.