Chapter 1.2, Problem 2E

Solution

Explain how to find a quadratic function’s maximum value or minimum value when the function is given in intercept form.

The intercept form of a quadratic equation is as follows:

  $f(x)=a(x-b)(x-c)$

We know that the x-intercepts pulled from the equation are b and c.

We also know that parabolas are symmetrical, that the mid-point between any two corresponding points (points opposite each other) in a parabola will help to tell us where the vertex is.

Using the following steps we can determine the vertex.

X-coordinate of the vertex $x=\frac{b+c}{2}=p$

X-coordinate of the vertex $=p$

Then we simply substitute this value back into the equation to find the y-coordinate of the vertex.

  $\begin{array}{l}f(p)=a(p-b)(p-c)\\ =q\end{array}$

Now we have a vertex $(p,q)$ .

Lastly we just need to check our leading coefficient a.

• If a is positive, then we have a minimum value of q.
• If a is negative, then we have a maximum value of q.

Concept Used:

• If the coefficient of $x^2$ that is ‘a’ is positive, the function has a minimum. 
• If $a>0$ minimum value (graph open up).
• If the coefficient of $x^2$ that is ‘a’ is negative, the function has a maximum.
• If $a<0$ maximum value (graph open down).