Answer
(a)
vertex: $(3, 4)$
x-intercepts: $1$ and $5$
y-intercept: $-5$
(b)
maximum value: $4$
(c)
domain: $(-\infty, +\infty)$
range: $-\infty, 4]$
Work Step by Step
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 $(-\infty, 4]$.
