Why is $(h,k)$ in the vertex formula of a parabola $=a(x-h)^2 +k$ the vertex?

Question

Why is it that in the form $y=a(x-h)^2+k$ of a parabola, that $(h,k)$ is the coordinate of the vertex? I am reviewing Algebra and cannot find a reasonable explanation anywhere.The highest level of math I am familiar is with Pre-Calculus. Thanks!

Answer 1

Consider the graph of the parabola $y=ax^2$. Its vertex is clearly at $(0,0)$. Now, if you replace $x$ with $x-h$ in any equation, its graph gets shifted to the right by a distance of $h$. Similarly, replacing $y$ by $y-k$ shifts the graph up by $k$. If we make both of these substitutions in the above equation of the parabola, its vertex gets shifted to $(h,k)$. The resulting equation after this substitution is $y-k=a(x-h)^2$, or $y=a(x-h)^2+k$.

Answer 2

Since $(x-h)^2$ is always nonnegative, for any value of $x$, it's at its smallest when the thing in parens is zero, i.e., when $x = h$. If $a$ is positive, this will be where the parabola has its minimum, i.e., $x = h$ is the $x$-coord of the vertex. (If $a$ is negative, this is the location of the max, but the same result applies: $x = h$ is the $x$-coord of the vertex).

At $x = h$, the $y$-value is $a(h-h)^2 + k$< which is just $k$, so the vertex is at location $(h, k).

Answer 3

Consider the situation using transformations of functions.

We know the vertex of $y=x^2$ is located at $(0,0)$.

To translate the parabola horizontally $h$ units, we modify the input ($x$) value like $y=(x-h)^2$. Thus, the vertex is now at $(h,0)$.

The $a$-value is a vertical stretch or compression, and also determines whether the parabola is reflected over the $x$-axis. In any of these cases, the vertex of the graph of $y=a(x-h)^2$ doesn't move, so the vertex is still $(h,0)$.

Finally, we can vertically shift the graph $h$ units by adjusting the function values, like so, $y=a(x-h)^2+k$, resulting in a vertex located at $(h,k)$.