Discussion of Quadratic Equations and Their Graphs

Polynomials are of the form y = b + c1x + c2x2 + c3x3 + c4x4 + … where b is the constant and c1, c2, c3, … are the coefficients of the x terms. A polynomial can have any number of terms. The highest power is the degree of the polynomial.

Quadratic equations are polynomials of degree 2. The general form of a quadratic equation is y = ax2 + bx + c. The simplest quadratic equation is y = x2. When quadratic equations are graphed, they form a parabola. The coefficient of the x2 term, a, helps determine the shape of the graph. Try graphing these equations on the TI-83:
y = x2; y = 3x2; y = 5x2; y = 0.5x2; y = 0.25x2.
You should see that the greater the value of a, the “steeper” the parabola is. The smaller the value of a, the more “shallow” the parabola is. All of these parabolas are said to “open up.” Now try graphing y = -4x2 and y = -0.75x2. Do you see that these parabolas “open down?” So the value of a also determines the direction the parabola opens.

The vertex of the parabola is where it turns. If the parabola opens up, the vertex is the lowest point. If it opens down, the vertex is the greatest point. All of the parabolas of the form y = ax2 have a vertex at (0,0). If the vertex of the parabola is not at the origin, the graph will be shifted or translated to the new position. Try graphing y1 = x2 + 3. This is the same shape as the graph y = x2, only moved 3 units up. If you rewrite the equation in the form y 3 = x2, you can see that when 3 is subtracted from y, the curve is moved up the relationship of moving up the y-axis 3 units. Now try graphing y1 = x2 and y2 = (x 8)2. You should see that the graph of y1 = x2 has been moved 8 units to the right.

In general, replacing y by yk in an equation translates the graph k units up, and replacing x by x - h in an equation translates the graph by h units to the right. So the vertex form of an equation for a parabola at vertex (h, k) is yk = a(xh)2.

To see how a quadratic equation of the form yk = a(xh)2 relates to one in general form, we can expand the first equation and rewrite it: y = a(x2 - 2xh + h2) + k so
y = ax2 2ahx + (ah2 + k). The general form is y = ax2 + bx + c, so a = a;b = -2ah and
c = ah2 + k. You can solve for h and k to find the vertex of the equation. and
k = cah2.