Graphing Quadratic Functions Using Vertex Form and Intercept Form
Reference

Parabola and Its Properties

Concept

Parabola

A parabola is a curve that is geometrically defined as the locus of all points equidistant from a line and a point not on the line. The line is called the directrix, and the point is the focus of the parabola. In the following applet, point P is equidistant from the directrix d and the focus F.
parabola

A parabola can be vertical or horizontal. A vertical parabola can open upward or downward. In comparison, a horizontal parabola can open to the left or the right. The graph of a quadratic function is a vertical parabola.

Vertical and horizontal parabolas
Concept

Direction of a Parabola

A parabola either opens upward or downward. This is the direction of the parabola . If the leading coefficient a of the corresponding equation is positive, the parabola opens upward. If the coefficient is negative, the parabola opens downward.

direction of a parabola
Concept

Vertex of a Parabola

Because a parabola either opens upward or downward, there is always one point that is the absolute maximum or absolute minimum of the function. This point is called the vertex.
vertex of a parabola showing maximum or minimum button animation
At the vertex, the function changes from increasing to decreasing, or vice versa.
Concept

Zeros of a Parabola

A parabola can intersect the x-axis at zero, one, or two points. Since the function's value at an x-intercept is always 0, these points are called zeros, or sometimes roots.
zeros of a parabola

Zeros of a parabola are sometimes referred to as zeros of a function. These are the x-coordinates of the points where a parabola intersects the x-axis.

Reference

Writing the Equation of a Parabola

The equation of a parabola can be derived from its focus and directrix by using the geometric properties of the parabola.

Downward opening parabola with its focus and directix shown
Given any point on the parabola, its distance to the focus and to the directrix are equal. Then, by using the distance formula, the lengths can be compared to get the equation for the parabola. Follow these steps for a more detailed explanation of the process.
1
Identify Focus and Directrix
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Determine the coordinates of the focus and the equation of the directrix. The focus of the parabola is a fixed point towards which the parabola curves. The directrix is a fixed line, perpendicular to the axis of symmetry of the parabola.

For this parabola, the coordinates of its focus are F(- 2, - 4.5), and its directrix is given by y=1.5.

2
Choose an Auxiliary Point
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Any point on a parabola has the same distance to the focus F as to the directrix. Choose an arbitrary point on the parabola to define these distances.

An auxiliary point P
3
Express the Distances
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Consider the two distances defined in Step 2. The distance between P and the directrix is the distance between P and the closest point on the line. Since these share the same x-coordinate, their distance is the difference between their y-values.

Distance between point P and the directix
Now, points P and F do not share any coordinates. However, their distance can be found using the distance formula. Substitute the coordinates of the points into the formula and simplify the resulting expression.
PF = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
PF = sqrt(( x-( -2))^2 + ( y-( -4.5))^2)
PF=sqrt((x+2)^2+(y+4.5)^2)
Here are the found lengths. PQ&=1.5-y PF&=sqrt((x+2)^2+(y+4.5)^2)
4
Compare Line Segments
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According to the definition of a parabola, point P is equidistant from points F and Q. In other words, the distances PD and PF are equal. 1.5-y=sqrt((x+2)^2+(y+4.5)^2) The equation for the parabola can now be found by solving this equation for y.
1.5-y=sqrt((x+2)^2+(y+4.5)^2)
(1.5-y)^2=(x+2)^2+(y+4.5)^2
1.5^2-3y+y^2=(x+2)^2+(y+4.5)^2
Since the equation is being solved for y, it is best to expand (y+4.5)^2 and keep the expression (x+2)^2 in its factored form.
1.5^2-3y+y^2=(x+2)^2+(y+4.5)^2
1.5^2-3y+y^2=(x+2)^2+y^2+9y+4.5^2
2.25-3y+y^2=(x+2)^2+y^2+9y+20.25
Solve for y
- 3y+y^2=(x+2)^2+y^2+9y+18
- 3y+y^2-y^2-9y=(x+2)^2+18
y^2-y^2+(- 3y)-9y=(x+2)^2+18
-12y=(x+2)^2+18
y=(x+2)^2+18/-12
Simplify right-hand side
y=- [(x+2)^2+18]/12
y=- (x+2)^2-18/12
y=- (x+2)^2/12-18/12
y=- (x+2)^2/12-1.5
Here is the equation for the parabola shown in the graph.
Downward opening parabola with its equation
Exercises