Solution Bank Quadratic Functions Target F
Answers to the Practice Problems are below in random order.
\(\begin{align}&D=9; x=\frac{2}{5}, 1\ & \ \ \ \ &y=(x+4)^2-3\ & \ \ \ \ &y=-0.0107x^2+4.3151x-304.014; \$105,000\\\\
&y=2(x-5)^2+3\ & \ \ \ \ &y=\frac{1}{5}(x-2)^2+3\ & \ \ \ \ &y=-3(x-8)^2+5\\\\
&y=-\frac{3}{2}(x+2)^2+7\ & \ \ \ \ &y=-\frac{1}{2}(x+4)(x-10)\ & \ \ \ \ &y=-\frac{7}{243}(x-27)^2+21\ \text{or}\ y=-\frac{7}{243}x(x-54)\\\\
&y=-3(x+5)(x+2)\ & \ \ \ \ &y=2(x-3)(x+1)\ & \ \ \ \ &y=\frac{4}{5}(x+6)(x-3)\\\\
&y=\frac{4}{3}x(x+9)\ & \ \ \ \ &x=-4\pm i\sqrt{3}\end{align}\)
\(\text{Vertex:}\ (2,-18);\ \text{Domain:}\ (-\infty,\infty);\ \text{Range:}\ [-18,\infty);\ \text{x-intercepts:}\ (-1,0)\ \text{and}\ (5,0)\)
Answers to the Practice Problems are below in random order.
\(\begin{align}&D=9; x=\frac{2}{5}, 1\ & \ \ \ \ &y=(x+4)^2-3\ & \ \ \ \ &y=-0.0107x^2+4.3151x-304.014; \$105,000\\\\
&y=2(x-5)^2+3\ & \ \ \ \ &y=\frac{1}{5}(x-2)^2+3\ & \ \ \ \ &y=-3(x-8)^2+5\\\\
&y=-\frac{3}{2}(x+2)^2+7\ & \ \ \ \ &y=-\frac{1}{2}(x+4)(x-10)\ & \ \ \ \ &y=-\frac{7}{243}(x-27)^2+21\ \text{or}\ y=-\frac{7}{243}x(x-54)\\\\
&y=-3(x+5)(x+2)\ & \ \ \ \ &y=2(x-3)(x+1)\ & \ \ \ \ &y=\frac{4}{5}(x+6)(x-3)\\\\
&y=\frac{4}{3}x(x+9)\ & \ \ \ \ &x=-4\pm i\sqrt{3}\end{align}\)
\(\text{Vertex:}\ (2,-18);\ \text{Domain:}\ (-\infty,\infty);\ \text{Range:}\ [-18,\infty);\ \text{x-intercepts:}\ (-1,0)\ \text{and}\ (5,0)\)