## Precalculus: Mathematics for Calculus, 7th Edition

Rewrite f(x) in standard form, $f(x)=a(x-h)^{2}+k$, read the vertex, (h,x) For the y-intercept, calculate 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. With this information (and possible additional points) sketch a graph Read the graph for range and domain. ------------------ a. $f(x)=-3x^{2}+6x-2 \qquad$factor out $-3$, $f(x)=-3(x^{2}-2x) -2 \quad$... complete the square $f(x)=-3(x^{2}-2(1)x+1^{2}-1^{2}) -2$ $f(x)=-3(x-1)^{2}+3-2$ $f(x)=-3(x-1)^{2}+1$ b. vertex: $(h,k)=(1, 1)$, a=$-3$, opens down, the vertex is a maximum y-intercept: f(0) = $-2$ x-intercepts: f(x)=0 $-3(x-1)^{2}+1=0$ $1=3(x-1)^{2}\qquad/\div 3$ $(x-1)^{2}=\displaystyle \frac{1}{3}\qquad/\sqrt{..}$ $x-1=\pm\sqrt{\frac{1}{3}}$ $x=1\pm\sqrt{\frac{1}{3}}$ c. see image (one pair of additional points, either side of the vertex). d. domain: all reals, $\mathbb{R}$ range: $(-\infty,1]$