How To Find The Vertex Of A Quadratic Equation 10 Steps

x=−b2a{\displaystyle x={\frac {-b}{2a}}} x=−(9)(2)(1){\displaystyle x={\frac {-(9)}{(2)(1)}}} x=−92{\displaystyle x={\frac {-9}{2}}}

y=x2+9x+18{\displaystyle y=x^{2}+9x+18} y=(−9)(2)2+9(−9)(2)+18{\displaystyle y={\frac {(-9)}{(2)}}^{2}+9{\frac {(-9)}{(2)}}+18} y=814−812+18{\displaystyle y={\frac {81}{4}}-{\frac {81}{2}}+18} y=814−1624+724{\displaystyle y={\frac {81}{4}}-{\frac {162}{4}}+{\frac {72}{4}}} y=(81−162+72)4{\displaystyle y={\frac {(81-162+72)}{4}}} y=−94{\displaystyle y={\frac {-9}{4}}}

x2+4x+1=0{\displaystyle x^{2}+4x+1=0} x2+4x+1−1=0−1{\displaystyle x^{2}+4x+1-1=0-1} x2+4x=−1{\displaystyle x^{2}+4x=-1}

x2+4x+1=0{\displaystyle x^{2}+4x+1=0} x2+4x+1−1=0−1{\displaystyle x^{2}+4x+1-1=0-1} x2+4x=−1{\displaystyle x^{2}+4x=-1}

x2+4x+1=0{\displaystyle x^{2}+4x+1=0} x2+4x+1−1=0−1{\displaystyle x^{2}+4x+1-1=0-1} x2+4x=−1{\displaystyle x^{2}+4x=-1}

x2+4x+1=0{\displaystyle x^{2}+4x+1=0} x2+4x+1−1=0−1{\displaystyle x^{2}+4x+1-1=0-1} x2+4x=−1{\displaystyle x^{2}+4x=-1}

(42)2=22=4{\displaystyle ({\frac {4}{2}})^{2}=2^{2}=4}. Now, add 4 to both sides of the equation to get the following: x2+4x+4=−1+4{\displaystyle x^{2}+4x+4=-1+4} x2+4x+4=3{\displaystyle x^{2}+4x+4=3}

(42)2=22=4{\displaystyle ({\frac {4}{2}})^{2}=2^{2}=4}. Now, add 4 to both sides of the equation to get the following: x2+4x+4=−1+4{\displaystyle x^{2}+4x+4=-1+4} x2+4x+4=3{\displaystyle x^{2}+4x+4=3}

x2+4x+4=−1+4{\displaystyle x^{2}+4x+4=-1+4} x2+4x+4=3{\displaystyle x^{2}+4x+4=3}