Introduction to Parabolas and Circles
The Cartesian-coordinate graph of a quadratic equation is a parabola . In Cartesian coordinates, a quadratic equation looks like this:
y = ax ^{2} + bx + c
where a, b , and c are real-number constants, and a ≠ 0. (If a = 0, then the equation is linear, not quadratic.) To plot a graph of an equation that appears in the above form, first determine the coordinates of the following point ( x _{0} , y _{0} ):
x _{0} = − b /(2 a )
y _{0} = c − b ^{2} /(4 a )
This point represents the base point of the parabola; that is, the point at which the curvature is sharpest, and at which the slope of a line tangent to the curve is zero. Once this point is known, find four more points by “plugging in” values of x somewhat greater than and less than x _{0} , and then determining the corresponding y -values. These x -values, call them x _{−2} , x _{−1} , x _{1} , and x _{2} , should be equally spaced on either side of x _{0} , such that:
x _{−2} < x _{−1} < x _{0} < x _{1} < x _{2}
x _{−1} − x _{−2} = x _{0} − x _{−1} = x _{1} − x _{0} = x _{2} − x _{1}
This will give five points that lie along the parabola, and that are symmetrical relative to the axis of the curve. The graph can then be inferred (that means we make an educated guess!) if the points are wisely chosen. Some trial and error might be required. If a > 0, the parabola opens upward. If a < 0, the parabola opens downward.
Let’s go ahead and try this with a concrete example.
Plotting A Parabola
Consider the following formula:
y = x ^{2} + 2 x + 1
Using the above formula to calculate the base point:
Therefore,
( x _{0} , y _{0} ) = (−1, 0)
Fig. 6-7 . Graph of the quadratic equation y = x ^{2} + 2 x + 1.
This point is plotted first, as shown in Fig. 6-7. Next, plot the points corresponding to x _{−2} , x _{−1} , x _{1} , and x _{2} , spaced at 1-unit intervals on either side of x _{0} , as follows:
x _{−2} = x _{0} − 2 = −3
y _{−2} = (−3) ^{2} + 2 × (−3) + 1 = 9−6 + 1 = 4
Therefore,
( x _{−2} , y _{−2} ) = (−3, 4)
x _{−1} = x _{0} − 1 = −2
y _{−1} = (−2) ^{2} + 2 × (−2) + 1 = 4 − 4 + 1 = 1
Therefore,
( x _{−1} , y _{−1} ) = (−2, 1)
x _{1} = x _{0} + 1 = 0
y _{1} = 0 ^{2} + 2 × 0 + 1 = 0 + 0 + 1 = 1
Therefore,
( x _{1} , y _{1} ) = (0,l)
x _{2} = x _{0} + 2 = 1
y _{2} = I ^{2} + 2 × 1 + 1 = 1 + 2 + 1 = 4
Therefore,
( x _{2} , y _{2} ) = (1, 4)
From these five points, the curve can be inferred.
Plotting Another Parabola
Let’s try another example, this time with a parabola that opens downward. Consider the following formula:
y = −2 x ^{2} + 4 x − 5
The base point is:
x _{0} = −4/−4 = 1
y _{0} = −5 − 16/(−8) = −5 + 2 = −3
Therefore,
( x _{0} , y _{0} = (1, −3)
This point is plotted first, as shown in Fig. 6-8.
Fig. 6-8 . Graph of the quadratic equation y = −2 x ^{2} + 4 x −5.
Next, plot the following points:
x _{−2} = x _{0} − 2 = −1
y _{−2} = −2 × (−1) ^{2} + 4 × (−1) − 5 = −2 − 4 − 5 = −11
Therefore,
( x _{−2} , y _{−2} ) = (−1, −11)
x _{−1} = x _{0} − 1 = 0
y _{−1} = −2 × 0 ^{2} + 4 × 0 − 5 = −5
Therefore,
( x _{−1} , y _{−1} ) = (0, − 5)
x _{1} = x _{0} + 1 = 2
y _{1} = −2 × 2 ^{2} + 4 × 2 − 5 = −8 + 8 − 5 = −5
Therefore,
( x _{1} , y _{1} ) = (2, − 5)
x _{2} = x _{0} + 2 = 3
y _{2} = −2 × 3 ^{2} + 4 × 3 − 5 = −18 + 12 − 5 = −11
Therefore,
( x _{2} , y _{2} ) = (3, −11)
From these five points, the curve can be inferred.
Equation of a Circle
The general form for the equation of a circle in the xy -plane is given by the following formula:
( x − x _{0} ) ^{2} + ( y − y _{0} ) ^{2} = r ^{2}
where ( x _{0} , y _{0} ) represents the coordinates of the center of the circle, and r represents the radius . This is illustrated in Fig. 6-9. In the special case where the circle is centered at the origin, the formula becomes:
x ^{2} + y ^{2} = r ^{2}
Fig. 6-9 . Circle centered at ( x _{0} , y _{0} ) with radius r .
Such a circle intersects the x axis at the points ( r , 0) and (− r , 0); it intersects the y axis at the points (0, r ) and (0, − r ). An even more specific case is the unit circle:
x ^{2} + y ^{2} = 1
This curve intersects the x axis at the points (1,0) and (−1, 0); it intersects the y axis at the points (0,1) and (0, −1).
Parabolas and Circles Practice Problems
PROBLEM 1
Draw a graph of the circle represented by the equation ( x − 1) ^{2} + ( y + 2) ^{2} = 9.
SOLUTION 1
zBased on the general formula for a circle, we can determine that the center point has coordinates x _{0} = 1 and y _{0} = −2. The radius is equal to the square root of 9, which is 3. The result is a circle whose center point is (1,−2) and whose radius is 3. This is shown in Fig. 6-10.
Fig. 6-10 . Illustration for Problem 6-3.
PROBLEM 2
Determine the equation of the circle shown in Fig. 6-11.
Fig. 6-11 . Illustration for Problem 6-4.
SOLUTION 2
First note that the center point is (−8, −7). That means x _{0} = −8 and y _{0} = −7. The radius, r , is equal to 20, so r ^{2} = 20 × 20 = 400. Plugging these numbers into the general formula gives us the equation of the circle, as follows:
( x − x _{0} ) ^{2} + ( y − y _{0} ) ^{2} = r ^{2}
[ x − (−8)] ^{2} + [ y − (−7)] ^{2} = 400
( x + 8) ^{2} + ( y + 7) ^{2} = 400
Practice problems for these concepts can be found at: The Cartesian Plane Practice Test.