Parabolas and Circles Help (page 2)

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:

• x ₀ = −2/2 = −1

  y ₀ = 1 − 4/4 = 1 − 1 = 0

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.