6.2: The Vertex and Parabola Features | Algebra 1 Honors

How it works

Vertex form reveals a parabola extremum and axis of symmetry. A positive leading coefficient opens upward and gives a minimum; a negative coefficient opens downward and gives a maximum. In a context, interpret the coordinates using the input and output units and restrict the domain appropriately.

y=a(x-h)^2+k\quad\Rightarrow\quad\text{vertex }(h,k)

Worked examples

Example 1

Find the vertex and range of $\displaystyle y=2(x-3)^2-4$ .

The vertex is $\displaystyle (3,-4)$ . Since $\displaystyle a=2>0$ , the graph opens upward and the range is $\displaystyle y\ge-4$ .

Example 2

Write $\displaystyle y=x^2-4x+1$ in vertex form.

Complete the square: $\displaystyle x^2-4x+1=(x-2)^2-3$ . The vertex is $\displaystyle (2,-3)$ .

Problems

Problem 6.2.1

Find the vertex and axis of $\displaystyle y=-(x+1)^2+5$ .

Show worked solution

The vertex is $\displaystyle (-1,5)$ and the axis is $\displaystyle x=-1$ . It opens downward, so the maximum is $\displaystyle 5$ .

Problem 6.2.2

Find the minimum of $\displaystyle y=x^2+6x+11$ .

Show worked solution

Complete the square: $\displaystyle y=(x+3)^2+2$ . The minimum is $\displaystyle 2$ at $\displaystyle x=-3$ .

Problem 6.2.3

Read the vertex, axis, and zeros of the graphed parabola $\displaystyle y=(x-1)^2-4$ .

An upward parabola has equation y = (x − 1) squared − 4. — Read the turning point, symmetry line, and horizontal-axis crossings.

Show worked solution

The vertex is $\displaystyle (1,-4)$ and the axis is $\displaystyle x=1$ . Setting $\displaystyle y=0$ gives $\displaystyle (x-1)^2=4$ , so the zeros are $\displaystyle x=-1$ and $\displaystyle x=3$ .

An upward parabola has equation y = (x − 1) squared − 4. Vertex (1, −4); axis x = 1; zeros −1 and 3. — Vertex (1, −4); axis x = 1; zeros −1 and 3.

Lesson Problems PDF Solutions PDF Unit MCQ

How it works

Worked examples

Example 1

Example 2

Problems

Problem 6.2.1

Problem 6.2.2

Problem 6.2.3

Need help with this topic?