Question

gerald graphs the function $f(x)=(x-3)^2 - 1$. which statements are true about the graph? select three options.
$square$ the domain is ${x|x \geq 3}$.
$square$ the range is ${y|y \geq -1}$.
$square$ the function decreases over the interval $(-\infty, 3)$.
$square$ the axis of symmetry is $x = -1$.
$square$ the vertex is $(3,-1)$.

Explanation:

Step1: Identify vertex form

The function is in vertex form $f(x) = (x-h)^2 + k$, where $(h,k)$ is the vertex. Here $h=3$, $k=-1$.

Step2: Determine domain

For quadratic functions, all real numbers are allowed for $x$. Domain: $(-\infty, \infty)$.

Step3: Determine range

Since the coefficient of $(x-3)^2$ is positive, the parabola opens upward. The minimum $y$-value is $k=-1$, so range is $y \geq -1$.

Step4: Find axis of symmetry

Axis of symmetry is $x=h$, so $x=3$.

Step5: Analyze increasing/decreasing

For upward-opening parabola, it decreases left of vertex: interval $(-\infty, 3)$.

Step6: Verify vertex

From vertex form, vertex is $(h,k)=(3,-1)$.

Answer:

• The range is $\{y| y \geq -1\}$.
• The function decreases over the interval $(-\infty, 3)$.
• The vertex is $(3,-1)$.