Video Transcript
This is a table for 𝑓 of 𝑥 equals
𝑥 squared plus two. Complete it by finding the values
of 𝑎, 𝑏, and 𝑐.
Remember to complete a table of
values for a function of the form 𝑓 of 𝑥 equals 𝑘𝑥 squared plus 𝑐, we
substitute each value of 𝑥 into the function. So, to find the value of 𝑎, we’ll
substitute 𝑥 equals negative two into the function 𝑓 of 𝑥 equals 𝑥 squared plus
two. In other words, 𝑎 is 𝑓 of
negative two. Since our function is 𝑥 squared
plus two, that’s negative two squared plus two. And of course, negative two squared
is four. So, this is four plus two, which is
equal to six. We now see that we’re not
interested in the value of the function when 𝑥 equals negative one. And we’re told that when 𝑥 is
equal to zero, 𝑓 of 𝑥 is two.
So, in order to find 𝑏, we’re
going to let 𝑥 be equal to one. This means 𝑏 is the value of the
function at this point; it’s 𝑓 of one. That’s one squared plus two, which
is one plus two, which, of course, equals three. Finally, we find the value of 𝑐 by
substituting 𝑥 equals two into our function. This means 𝑐 is equal to 𝑓 of
two, which is two squared plus two. Once again, that’s four plus two,
which is equal to six.
Let’s check our method by
calculating 𝑓 of zero and checking that it gives the correct output of two in the
table. 𝑓 of zero is zero squared plus
two. That’s zero plus two or two as we
expected. Since this value of 𝑓 of zero
corresponds to the value given in our table, we can be fairly confident in our
method. So, 𝑎 is equal to six, 𝑏 is equal
to three, and 𝑐 is equal to six.
Now, in this example, we could have
also calculated the value of 𝑓 of negative one. We substitute 𝑥 equals negative
one into the function, and we get negative one squared plus two, which is equal to
three. Adding these values to our table
and we now might notice that there’s a symmetry to our values of 𝑓 of 𝑥. This is not accidental. The graphs of quadratic functions
are symmetrical about a vertical line, as demonstrated in the diagram. For very simple quadratic
functions, as in the one in this question, this can be observed in the table of
values and that gives us a helpful way of checking our results.
We also notice that whilst there’s
reflectional symmetry between the coordinates generated, the values of the function
don’t increase linearly. And this means that we have to join
the coordinates with a smooth curve instead of a straight line.
