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Brush up on your algebra For all of these tutorials on nonlinear functions, you should be familiar with the
algebra of exponents. Also, you should review
solving quadratic equations by factoring and by using the quadratic formula.
Fundamentals
The relationship between two quantities is often best modeled by a curved line rather than a straight line. The simplest function whose graph that is not straight line is a
quadratic function.
Notice that, in the example shown, the curves do not cross in the interval $[a, b]$. We focus on this case first.
Quadratic function
A quadratic function of the variable $x$ is a function that can be written in the form
$f(x) = ax^2 + bx + c$ \gap[40] \t
where $a, b,$ and $c$ are fixed numbers (with $a \neq 0$).
Examples
1. | $f(x) = 3x^2-2x+1$ | |
2. | $\displaystyle g(x) = -\frac{x^2}{2}$ | |
3. | $g(x) = 3x+1$ | |
4. | $\displaystyle h(x) = %11$ | | |
Graph of a quadratic function
The graph of a quadratic function is a
parabola (see the figure below).
Concavity: If the coefficient $a$ of $x^2$ is positive, it is
concave up (as in the figure below when you press "$a \gt 0$"). If $a$ is negative, it is
concave down (as in the figure below when you press "$a \lt 0$").
Vertex: The
vertex of this parabola (press the "Vertex" button above) occurs at the point on the graph with
Vertex: $\displaystyle x = -\frac{b}{2a} \qquad $
y-Intercept: (Press the "$y$-intercept" button above) It crosses the $y$-axis at
x-Intercept(s): (Press the "$x$-intercept" button above) It crosses the $x$-axis at the solution(s) of the quadratic equation $ax^2+bx+c = 0$ (if there are any). We can sometimes solve this equation by factoring the left-hand side. If it does not factor, we can always use the quadratic formula:
Note that, if the discriminant $b^2-4ac$ is negative, then there are no $x$-intercepts as then the quadratic equation has no real solutions.
Symmetry: It is symmetric around the vertical line through the vertex.
Example
%Let $f(x) = -3x^2-6x-3.$ %Therefore $a = -3, \ b = -6,\ c = -3.$
Concavity: As $a=-3$ is negative, the graph is concave down (press the "a < 0" button above to see its shape).
Vertex: (press the "Vertex" button above) The $x$-coordinate is
$x = -\dfrac{b}{2a} = -\dfrac{(-6)}{2(-3)} = -1$
The corresponding $y$-coordinate is
$y = f(-1) = -3(-1)^2-6(-1)-3 = -3+6-3 = 0$.
So, the vertex is at $(-1,0)$:
y-Intercept: It crosses the $y$-axis at $y = c = -3$.
x-Intercept(s): It crosses the $x$-axis at the solution(s) of $ax^2+bx+c = 0$:
$-3x^2-6x-3 = 0$
\\ $\implies -3(x^2+2x+1) = 0$
\\ $\implies -3(x+1)^2 = 0$ \gap[4] Luckily, the left-hand side factors.
\\ $\implies x = -1$. \gap[4] Only one $x$-intercept, as already seen on the graph.
Consider $f(x) = %15$.
Fill in the coordinates of the following points.
Now use this information to get the graph for $y = f(x)$: Drag the vertex and the second point in the graph below into position to adjust it.
Applications
The population of Roman Catholic nuns in the U.S. during the last 25 years of the last century can be modeled by
$P(t) = %30$ thousand nuns $\qquad (5 \leq t \leq 25)$,
where $t = 0$ represents January 1970 (so that January 1975 is represented by $t = 5$).
Recall that the
revenue resulting from one or more transactions is the total payment received. (See the
tutorial on functions and models.) Thus, if $q$ units of some item are sold at $p$ dollars per unit, the revenue resulting from the sale is
Revenue = Price $\times$ Quantity
$R = pq$.
The
Trusted Genuine Masters company produces forged paintings for sale at well-known auction houses. The company estimates that its demand equation is
paintings sold per month when priced at $p$ dollars each. It has monthly fixed costs of &D&%41 and variable costs of &D&%42 per painting.
a.
Find the company's monthly revenue, cost, and profit as functions of $p$, the price per painting.
Now try the exercises in %4, some the %8, or move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.
Last Updated: September, 2017
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