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\(\newcommand{\highlight}[1]{{\color{magenta}{{#1}}}}\newcommand{\lighthigh}[1]{{\color{blue}{{#1}}}}\)

Find the vertex of each quadratic function.

$r(x)=-2(x+4)^2+10$

$s(x)=5(x-1)^2+2$

$t(x)=(x-10)^2-5$

$u(x)=3(x+7)^2-13$

Solution

The vertex of $r(x)=-2(x\highlight{{}+4})^2\lighthigh{{}+10}$ is $(\highlight{-4},\lighthigh{10})\text{.}$

The vertex of $s(x)=5(x\highlight{{}-1})^2\lighthigh{{}+2}$ is $(\highlight{1},\lighthigh{2})\text{.}$

The vertex of $t(x)=(x\highlight{{}-10})^2\lighthigh{{}-5}$ is $(\highlight{10},\lighthigh{-5})\text{.}$

The vertex of $u(x)=3(x\highlight{{}+7})^2\lighthigh{{}-13}$ is $(\highlight{-7},\lighthigh{-13})\text{.}$

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