## College Algebra 7th Edition

(a) vertex: $(3, 4)$ x-intercepts: $1$ and $5$ y-intercept: $-5$ (b) maximum value: $4$ (c) domain: $(-\infty, +\infty)$ range: $-\infty, 4]$
RECALL: The standard form of a quadratic function is $f(x) = a(x-h)^2+k$ where $(h, k)$ is the vertex. (a) The parabola opens downward so the vertex is the maximum point of the graph. Thus, the vertex is $(3, 4)$ This means that the tentative equation of the given function is: $f(x) = a(x-3)^2+4$ To find the value of $a$, substitute the coordinates of any point on the parabola. Using the point $(1, 0)$ gives: $f(x) = a(x-3)^2+4 \\0=a(1-3)^2+4 \\0=a(-2)^2+4 \\0=a(4)+4 \\-4 = 4a \\\dfrac{-4}{4} = \dfrac{4a}{4} \\-1=a$ Thus, the function is $f(x) = -(x-3)^2+4$ The graph clearly shows that the x-intercepts are $1$ and $5$. To find the y-intercept, set $x=0$ then solve for $y$: $f(x) = -(x-3)^2+4 \\f(0) = -(0-3)^2+4 \\f(0) = -(-3)^2+4 \\f(0) = -9+4 \\f(0)=-5$ Thus, the y-intercept is $-5$. (b) The parabola opens downward so the function has a maximum value, which is the y-coordinate of the vertex. Thus, the maximum value if $f$ is $4$. (c) The domain of a quadratic function is the set of all real numbers, $(-\infty, +\infty)$ The y-values of the function are from $4$ and below. Thus, the range is $(-\infty, 4]$.