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Question: How do you find the vertex of the parabola\(y=3{{x}^{2}}+5x+8\)?...

How do you find the vertex of the parabolay=3x2+5x+8y=3{{x}^{2}}+5x+8?

Explanation

Solution

The above given equation y=3x2+5x+8y=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=ax2+bx+cy=a{{x}^{2}}+bx+c.

Complete step by step solution:
The given equation is:
⇒y=3x2+5x+8\Rightarrow y=3{{x}^{2}}+5x+8
Here we have to find the vertex of the parabola. The standard parabolic equation is y=ax2+bx+cy=a{{x}^{2}}+bx+c and to find the value of yy first we have to find the x-coordinate of the equation. We know the formula to find the x-coordinate which is −b2a\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,
⇒a=3,b=5,c=8\Rightarrow a=3,b=5,c=8
Now put the above values in the formula of x-coordinate, we get
⇒x=−52(3)\Rightarrow x=\dfrac{-5}{2\left( 3 \right)}
By more simplifying we get,
⇒x=−56\Rightarrow x=\dfrac{-5}{6}
Now we will put this value of x in the given equation we get,
⇒y=3(−56)2+5(−56)+8 ⇒y=3(2536)−256+8 \begin{aligned} & \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{aligned}
More solving it then we get
⇒y=2512−256+8 ⇒y=25−50+9612 ⇒y=7112 \begin{aligned} & \Rightarrow y=\dfrac{25}{12}-\dfrac{25}{6}+8 \\\ & \Rightarrow y=\dfrac{25-50+96}{12} \\\ & \Rightarrow y=\dfrac{71}{12} \\\ \end{aligned}
Hence we get the vertex of the parabola which is (−56,7112)\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:
⇒y=3x2+5x+8\Rightarrow y=3{{x}^{2}}+5x+8
Now we will simplify the above equation as:
⇒y=3(x2+53x)+8\Rightarrow y=3\left( {{x}^{2}}+\dfrac{5}{3}x \right)+8
Now add and subtract (56)2{{\left( \dfrac{5}{6} \right)}^{2}} in the above equation we get,
⇒y=3(x2+2(56)x+(56)2)−3(56)2+8 ⇒y=3(x+56)2+7112 \begin{aligned} & \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{aligned}
Now add and subtract 7112\dfrac{71}{12} in both side of the equation, we get,
⇒y−7112=3(x+56)2 ⇒13(y−7112)=(x+56)2 \begin{aligned} & \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{aligned}
Now we know the above equation is looking like an upward parabola equation:
⇒(x−x1)=4a(y−y1)\Rightarrow \left( x-{{x}_{1}} \right)=4a\left( y-{{y}_{1}} \right)
And which has vertex as:
⇒(x−x1)=0,(y−y1)=0\Rightarrow \left( x-{{x}_{1}} \right)=0,\left( y-{{y}_{1}} \right)=0
By comparing the above equation with13(y−7112)=(x+56)2\dfrac{1}{3}\left( y-\dfrac{71}{12} \right)={{\left( x+\dfrac{5}{6} \right)}^{2}}, we get
⇒(x+56=0,y−7112=0) ⇒(x=−56,y=7112) \begin{aligned} & \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{aligned}
Hence we get the same vertex as we get above(−56,7112)\left( \dfrac{-5}{6},\dfrac{71}{12} \right).