## 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. factor out $-1$, $f(x)=-(x^{2}-6x) +4 \quad$... complete the square $f(x)=-(x^{2}-2(3)x+3^{2}-3^{2})+4$ $f(x)=-(x-3)^{2}+9+4$ $f(x)=-(x-3)^{2}+13$ b. vertex: $(h,k)=(3, 13)$, a=$-1$, opens down, the vertex is a maximum y-intercept: f(0) = $4$ x-intercepts: f(x)=0 $-(x-3)^{2}+13=0$ $(x-3)^{2}=13\qquad/\sqrt{..}$ $x-3=\pm\sqrt{13}$ $x=3\pm\sqrt{13}$ c. see image (one pair of additional points, either side of the vertex). d. domain: all reals, $\mathbb{R}$ range: $(-\infty,13]$