A quadratic function is a function f(x) given by a
formula
of the form
f(x) = ax2 + bx + c a ≠ 0.
Such a function is characterized graphically as a parabola.
The three coefficients a , b, c in the above formula
de termine the
orientation , location, and relative breadth of the graph.
Important Features of the Graph.
• The sign of a determines the “orientation”, opening up or down.
• The vertex of the parabola is the (unique) highest (if a < 0)
or lowest (if a > 0) point on the graph.
• The x- coordinate of the vertex is
and y-coordinate of the vertex is the maximum or minimum
value of the function .
• The vertical line through the vertex is the axis; the parabola is
left-right symmetric about its axis.
• The y- intercept is (0, c) with y-coordinate c.
• The graph need not intersect the x-axis, but if it does, it occurs
at the x-intercepts
and it has such x-intercepts if and only if the
discriminant
b2 − 4ac ≥ 0.
Thus, given the formula for a quadratic function it is
quite easy
to sketch its graph. One just finds the vertex, the intercepts (when
they exist) a couple of points on one side of the axis, and then, using
the symmetry of the graph, just sketch in the parabola through the
plotted points .
Example 1. Let’s graph the function
f(x) = x2 − 5x + 6 = (x − 2)(x − 3).
Well,
• x = −b/2a = 5/2 is the axis and (5/2, f(5/2)) = (5/2,−1/4)
the vertex;
• The graph opens upward and has minimum value −1/4;
• The y-intercept is at y = 6 an the x-intercepts are at x = 2, 3.
Example 2. Let’s graph the function g(x) = −x2 − 1.
• The axis is x = 0, the y-axis, and the vertex is (0,−1);
• The graph opens downward and has a minimum value y = −1;
• Since b2 − 4ac = −4, the graph does not meet the x-axis.
So now we can sketch both graphs.
Some applications.
Say that in our hot-dog empire we have determined the demand and
cost functions
q = D(p) (demand) and cost in $ = C(q)
in terms of the price p and the demand q. Then the revenue , “price
times quantity”, and the profit, “revenue minus cost” are given by
R(p) = pq = pD(p) (revenue)
and
P(p) = R(p) − C(D(p)) = pq − C(D(p)) (profit).
Note that since the cost is given as a function C(q) of q, we have to
substitute the value of q in terms of the price p, that is, q = D(p).
Example. Suppose that the demand function is
linear, say
q = D(p) = −400p + 1400
where p is measured in dollars and q in hot-dogs sold per day. Then
the revenue, as a function of price, is
R(p) = pq = −400p2 + 1400p,
a quadratic function in p with graph
So the revenue is maximum when p = $1.75 and q = 700 and
revenue is $1225 per day.
Suppose next that the cost of each hot-dog is $0.50 and
the daily
fixed costs are $500. Then the cost function in terms of p is
C(D(p)) = 0.5q +500 = 0.5(−400p+1400)+500 = −200p+1200,
and the profit function is
So the profit is maximum when p = $2.00, the demand is q = 600,
revenue is R = $1200, cost is C = $800, and the profit P = $400.
Laws of Exponents. Recall that for b, c > 0 and all
x, y
An exponential function is a function f given by a
formula
f(x) = Abx where b > 0
The graph of such a function has roughly the form (for b ≠ 1)
You will note that the y-intercept of the graph of f(x) =
Abx is
at y = A and b determines the change Δy of the variable y = f(x)
for a change Δx of the independent variable x. So these are roles
similar to those of the slope and intercept of a linear function .
Example. Let f(x) be a linear function and g(x) be
an exponential
function, and suppose they satisfy
Well, we have that f and g satisfy
f(x) = mx + c and g(x) = Abx
[Note: We used a different letter c for the y-intercept of f(x) to avoid
confusion with the base b of g(x).] So we know that
c = A = 3
the y-intercepts of both functions. But then since f(1) = g(1) = 6,
we have that
f(1) = m ยท 1 + 3 = 6 and g(1) = 3b1 = 6,
so that m = 3 and b = 2. Thus,
f(x) = 3x + 3 and g(x) = 3(2)x.
Example. Suppose that y = f(x) is exponential with f(1) = 12
and f(3) = 48. We want to find A and b so that f(x) = Abx. But
so b = 2. But then f(1) = Ab1 = 2A = 12, so A =
12/2 = 6, and
finally,
f(x) = 6(2)x.
