Math101 · Independent Ontario learning library
Math101learn.math101.caVertex
The vertex of a parabola is its turning point and lies on the axis of symmetry. For $y=a(x-h)^2+k$, the vertex is $(h,k)$.
The vertex locates extreme values, organizes graphing, and answers optimization questions in motion, area, revenue, and design.
Intuition and core definition
The vertex of a parabola is its turning point and lies on the axis of symmetry. For $y=a(x-h)^2+k$, the vertex is $(h,k)$. If $a>0$, it is a minimum; if $a<0$, it is a maximum. The vertex is a point, whereas the axis is the line $x=h$.
Notation, language, and conditions
For standard form $y=ax^2+bx+c$, $h=-b/(2a)$ and $k=f(h)$. Completing the square converts standard to vertex form. In an application, the vertex’s coordinates carry input and output units and must be checked against the model’s domain.
Why this idea matters
The vertex is a parabola's turning point and packages its extremum value with the axis of symmetry.
A dependable method
- Identify the quadratic’s form and coefficient $a$.
- Read $(h,k)$ from vertex form, or compute $h=-b/(2a)$ in standard form.
- Evaluate the function at $h$ to find $k$.
- Classify the point as maximum or minimum using the sign of $a$.
- Check symmetry with inputs $h-d$ and $h+d$ and interpret the coordinates.
Worked example
Representations and interpretation
On a graph, the vertex is where the parabola changes direction. In a table centred at $h$, paired equal outputs surround the vertex row. Vertex form encodes horizontal and vertical translations directly.
Reasoning about variations
A model may restrict the domain so the mathematical vertex is not the actual maximum or minimum on the permitted interval. Optimization must compare the vertex with endpoints whenever the domain is bounded.
Common mistakes
How to check your work
- Substitute $h$ into the original function.
- Evaluate symmetric inputs around $h$.
- Convert to vertex form and verify $(h,k)$.
Practice
- Find the vertex of $y=(x-4)^2-7$.
- Find the vertex of $y=x^2+6x+1$.
- Does $y=-3(x+2)^2+5$ have a maximum or minimum?
Answers and brief solutions
Show answers
- $(4,-7)$ Vertex form gives $h=4$ and $k=-7$.
- $(-3,-8)$ $h=-6/2=-3$ and $f(-3)=9-18+1=-8$.
- Maximum $a=-3<0$, so the parabola opens downward.
Synthesis and transfer
For a projectile-height model, the vertex identifies maximum height and its time; substituting nearby symmetric times checks the calculation geometrically.
For $h(t)=-4.9t^2+19.6t+1$, the vertex occurs at $t=-19.6/[2(-4.9)]=2$ seconds. Substitution gives the maximum height, while the negative leading coefficient confirms the graph opens downward. Writing the model as $-4.9(t-2)^2+k$ makes horizontal symmetry about $t=2$ explicit. Times one second before and after the vertex must have equal heights, offering a numerical check. In an unrestricted quadratic the vertex always exists, but a contextual domain may exclude it; an optimization conclusion must verify that the vertex lies within the physically allowed interval.
Related topics
Teaching and accessibility note
Explore the idea
Function transformation
Change one quantity at a time and connect what moves to Vertex.
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Find the vertex of $y=(x-4)^2-7$.
- Vertex form gives $h=4$ and $k=-7$.
End of lesson
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