What is the vertex of the quadratic function given by the equationf(x)=-(x+3)^2+5

asked Mar 26, 2019 95.7k views

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$\text{The vertex formula of}\ f(x)=ax^2+bx+c:\\\\f(x)=a(x-h)^2+k\\\\(h,\ k)-vertex$

$\text{We have}\ f(x)=-(x+3)^2+5=-(x-(-3))^2+5$

$\text{Therefore your answer is}\ (-3,\ 5)$

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