## Precalculus: Mathematics for Calculus, 7th Edition

a. vertex: $(h,k)=(1, -3)$ y: intercept: $-1$ x-intercepts: $1\pm\sqrt{\frac{3}{2}}$ b. f has a minimum value at the vertex, $f(-2)=-3$ c. domain: all reals, $\mathbb{R}$ range: $[-3,\infty)$
Rewrite f(x) in standard form, $f(x)=a(x-h)^{2}+k$, read the vertex, (h,x) For the x-intercept, calcucate f(0) For the y- intercept, solve f(x) = 0 for x. If $a>0$, parabola opens up, the vertex is a minimum point, If $a<0$, parabola opens down, the vertex is a maximum Read the graph for range and domain. ------------------ a. $f(x)=2x^{2}-4x-1 \quad$... factor out $2$ $f(x)=2(x^{2}-2x) -1 \quad$... complete the square ... $=2(x^{2}-2x+1-1)-1$ $=2(x^{2}-2x+1)-2-1$ $=2(x-1)^{2}-3$ vertex: $(h,k)=(1, -3)$ x-intercept: f(0) = $-1$ y-intercepts: f(x)=0 $2(x-1)^{2}-3=0$ $2(x-1)^{2}=3$ $(x-1)^{2}=\displaystyle \frac{3}{2}\qquad/\sqrt{...}$ $x-1=\pm\sqrt{\frac{3}{2}}$ $x=1\pm\sqrt{\frac{3}{2}}$ b. f has a minimum value at the vertex, $f(1)=-3$ c. domain: all reals, $\mathbb{R}$ range: $[-3,\infty)$