The tutor shows an example of a typical problem involving a quadratic function.

Consider the following problem:

A quadratic function has vertex (-7,4) and passes through (-4,10). Find its equation.

Solution:

Since we are given the vertex, the useful form of the equation to consider is vertex form:

y=a(x-p)^2+q

in which the vertex is (p,q)

For our case, we insert (-7,4) into the equation:

y=a(x-(-7))^2+4

which becomes

y=a(x+7)^2+4

Now, we just need to find the value for a. To do so, we plug in the other given point (-4,10):

10=a(-4+7)^2+4

10=a(3)^2+4

10=9a+4

We subtract 4 from both sides:

6=9a

Next, we divide both sides by 9:

6/9=a

2/3=a

Our full equation, and the solution to the problem posed, is

y=(2/3)(x+7)^2+4

Source:

Travers, Kenneth et al. Using Advanced Algebra. Toronto: Doubleday, 1977.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.