Activity

1. The graph that represents $p(x)=(x-8{)}^2+1$ has its vertex (of a graph) at $(8,1)$. Here is one way to show, without graphing, that $(8,1)$ corresponds to the minimum value of $p$.

• When $x=8$, the value of $(x-8{)}^2$ is 0 because $(8-8{)}^2=0^2=0$.
• Squaring any number always results in a positive number, so when $x$ is any value other than 8, $(x-8)$ will be a number other than 0, and when squared, $(x-8{)}^2$ will be positive.
• Any positive number is greater than 0, so when $x\ne 8$, the value of $(x-8{)}^2$ will be greater than when $x=8$. In other words, $p$ has the least value when $x=8$.

The greatest value of a function is the maximum value. The lowest value of a function is the minimum value. Since infinity cannot be reached or touched, it is not considered a maximum or minimum.

In quadratic functions, the maximum or minimum value occurs at the vertex of the parabola.

2. Here are some quadratic functions and the coordinates of the vertex of the graph of each. Determine if the vertex corresponds to the maximum or the minimum value of the function.

Are you ready for more?

Extending Your Thinking

1.

Here is a portion of the graph of function $q$, defined by $q(x)=-x^2+14x-40$.

$ABCD$ is a rectangle. Points $A$ and $B$ coincide with the $x$-intercepts of the graph, and segment $CD$ just touches the vertex of the graph.

Find the area of $ABCD$. Show your reasoning.

Compare your answer:

54 square units.

For example:

• The expression can be rewritten in factored form as $(-x+10)(x-4)$, so the $x$-intercepts are $(4,0)$ and $(10,0)$, which means the length of $AB$ is 6 units. The $x$-coordinate of the vertex is halfway between 4 and 10, which is 7. Substituting 7 for $x$ in $(-x+10)(x-4)$ gives $3·3$ or 9, so the height of the rectangle is 9. The area is $6·9$ or 54 square units.
• The equation $q(x)=-x^2+14x-40$ can be written in vertex form as $q(x)=-1(x-7{)}^2+9$, so its vertex is at $(7,9)$ and the height of the rectangle is 9 units. Completing the square for $-x^2+14x-40=0$ gives the solutions $x=4$ and $x=10$, which correspond to the $x$-intercepts at points $A$ and $B$. This means the width of the rectangle is $10-4$ or 6 units.

Additional Resources

Identify a Maximum or Minimum

Recall that the maximum is the highest point of the graph and the minimum is the lowest point of the graph. A quadratic function shape is called a parabola, and a parabola will only have a maximum or a minimum.

Example 1

Graph $f(x)=(x-2{)}^2-1$.

In the vertex form (of a quadratic expression), $f(x)=a(x-h{)}^2+k$, tthe vertex (of a graph) is located at $(h,k)$.

For the function graphed, the vertex is at $(2,-1)$.

Since the leading coefficient, $a$, is positive, the parabola opens up. This means the vertex is a minimum.

Example 2

Graph $f(x)=-(x+1{)}^2+3$.

The vertex is located at $(-1,3)$.

Since the leading coefficient is negative, the parabola opens down and the vertex is a maximum.

Try it

Identify a Maximum or Minimum

For the function $f(x)=(x+4{)}^2-2$, identify the vertex and tell if it is a maximum or minimum.

Here is how to identify if a vertex is a maximum or minimum:

Step 1 - Identify the vertex $(h,k)$.

$(-4,-2)$

Step 2 - Identify if the parabola opens up or down. Since the leading coefficient is positive, the parabola opens up.

Step 3 - Identify if the vertex is a maximum or minimum. Since the parabola opens up, the vertex is a minimum.