## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$f(x)=\displaystyle \frac{3}{5}(x-4)^{2}-7$
$f(x)=\displaystyle \frac{3}{5}x^{2}$ In order to have a vertex in the point $(4,-7)$, this function needs to be moved $4$ units to the right and $7$ units down. $f(x)=a(x-h)^{2}+k$ Substitute $a=\displaystyle \frac{3}{5},h=4,k=-7$ $f(x)=\displaystyle \frac{3}{5}(x-4)^{2}-7$