The function is written as $f(x) = (x - 3)^2 - 5$. Compare this to $f(x) = (x - h)^2 + k$. What are $h$ and $k$ here?

For a parabola that opens upward, does the vertex represent a minimum or a maximum value of the function? Which coordinate of the vertex is that value? Frоm аnікο.​аi

In Desmos, type y = (x - 3)^2 - 5 to graph the function.

Tap or click on the lowest point of the parabola (its vertex). Desmos will show the coordinates of this point; the y-coordinate is the minimum value of $f(x)$. Aniкo A‍І ‌Тu⁠tor

The function is given as $f(x) = (x - 3)^2 - 5$. This is a quadratic function written in vertex form:

where the vertex of the parabola is at $(h, k)$.

Compare $f(x) = (x - 3)^2 - 5$ with $f(x) = (x - h)^2 + k$:

• $h = 3$
• $k = -5$

So the vertex is at the point $(3, -5)$.

The coefficient of the squared term $(x - 3)^2$ is positive (it is $+1$), which means the parabola opens upward. For an upward-opening parabola, the vertex is the minimum point on the graph. So the minimum value of $f(x)$ is the y-value of the vertex.

The y-value of the vertex is $-5$. Therefore, the minimum value of $f(x)$ is $-5$. Prepа‌red ‍bу​ Anіkο‌.а‍i