## Precalculus: Mathematics for Calculus, 7th Edition

a. vertex: $(h,k)=(-1, -4)$ y-intercept: $-1$ x-intercepts: $-1\displaystyle \pm\frac{2\sqrt{3}}{3}$ b. $a>0$, f has a minimum value at the vertex, $f(-1)=-4$ c. domain: all reals, $\mathbb{R}$ range: $[-4,\infty)$
Rewrite f(x) in standard form, $f(x)=a(x-h)^{2}+k$, read the vertex, (h,x) For the y-intercept, calcucate f(0) For the x- 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. $f(x)=3x^{2}+6x-1 \quad$... factor out $3$ $f(x)=3(x^{2}+2x) -1 \quad$... complete the square $=3(x^{2}+2x+1-1)-1$ $=3(x^{2}+2x+1)-3-1$ $=3(x+1)^{2}-4$ vertex: $(h,k)=(-1, -4)$ y-intercept: f(0) = $-1$ x-intercepts: f(x)=0 $3(x+1)^{2}-4=0$ $3(x+1)^{2}=4$ $(x+1)^{2}=\displaystyle \frac{4}{3}\qquad/\sqrt{...}$ $x+1=\pm\sqrt{\frac{4}{3}}$ $x=-1\displaystyle \pm\frac{2}{\sqrt{3}}=-1\pm\frac{2\sqrt{3}}{3}$ b. $a>0$, f has a minimum value at the vertex, $f(-1)=-4$ c. domain: all reals, $\mathbb{R}$ range: $[-4,\infty)$