Expert: Josh - 9/16/2007

Question
Hi
I'm having trouble figuring out how to calculate the formula for a parabola from a graph.
I would appreciate any help you can give me.
thanks

Answer
Hi Paul,

This can be done in a number of ways. Here are some of the things to look out for.

a) Concavity: A parabola is said to be convex-up if it looks like a cup "U"; convex-down if it looks like an umbrella "^".

b) Axis of symmetry: i.e., whether it is symmetrical about the x-axis or y-axis (after appropriate translation to account for certain offset).

c) Turning point: the coordinate where a parabola attains its maximum or minimum value.

Example 1: Consider an upright parabola. In the simplest case, its lowest point is at the origin (x,y)=(0,0) and it is symmetrical about the vertical (y) axis. This parabola is described by the equation y=x^2. [x-square is typed as x^2]

Example 2: [Variations]
a) Translation: To shift the parabola to the left by 1 unit, the equation changes to y=(x+1)^2. To shift the parabola to the right by 1 unit, the equation changes to y=(x-1)^2. For the second case, verify that the minimum point occurs at x=1,y=0. Generalization: To translate the parabola to the right by "c" units along the x-axis, we write y=(x-c)^2.

To shift the parabola vertically by "a" units, we write y=x^2+a. If a is positive, it is shifted up; if negative, it is shifted down. Of course, we can have a combination of shifts - in both the horizontal and vertical direction. Consider y=(x-c)^2+a, where c=2,a=3. We have a "U"-shape parabola, symmetrical about x=2 (i.e., shifted to the right by two units) and lifted up by 3 units along the vertical direction. Its minimum point occurs at x=2. To see this, note that "a" is simply an offset. Since the square of "x-c" is always positive, to minimize the value of "y", we must minimize (x-c). The best we can do is set x-c=0, which yields x=c. Thus, the turning point is (x,y)=(c,a).

Example 3: [Reflection about the x/y axis]. Given a parabola y=(x+c)^2, we introduce a minus sign to flip this curve vertically (reflecting it about the x-axis). We get y=-(x+c)^2.

Exercise: Given a "U"-shape parabola with a minimum point at x=1,y=4, write an equation for its reflection about the x-axis.
Solution: Draw a sketch to see this. The equation for the given parabola is y=(x-1)^2+4. Its reflection is (after introducing a minus sign) y=-[(x-1)^2+4], or y=-(x-1)^2-4.

Example 4: [Parabola on its side] A parabola symmetrical about the x-axis, with a vertex at (0,0) is given by x=y^2. Notice the similarity with y=x^2. In fact, the algebraic expression is the same, except we have swapped "x" with "y". So, everything mentioned before holds. To shift this parabola to the right by "d" units, we get (x-d)=y^2. Of course, we can take the square root on both sides to write this in the "standard" form, y=+sqrt(x-d) and y=-sqrt(x-d). The former describes the ascending part "/", the latter describes the descending part "\".

Exercise: Sketch the parabola (x-4)=(y-1)^2.
Solution: This is in the same form as x=y^2. We are looking at a parabola on its side. The parabola is "C"-shape, with a minimum point at x=4,y=1. It is symmetrical about y=1, so it is lifted up by 1 unit in the vertical direction.

Exercise: Write an equation for a ")"-shape parabola, symmetrical about y=-3, with a maximum point coinciding with x=0.5. Hence, find the turning point.
Solution: We already know a "C"-shape parabola has the form x=y^2. To flip this around about the vertical (y) axis, we introduce a minus sign to get -x=y^2. Interpretation: The parabola grows in the negative-x direction. Next, we account for the offsets. Since it is symmetrical about y=-3, we modify the equation to -x=(y+3)^2+c, where "c" is some constant (fixed value). Important: To find "c", we use the last piece of information, forcing the parabola to pass through x=0.5. Because the line x=0.5 passes through the parabola at the turning point, it must intersect intersect the axis of symmetry y=-3 and satisfy the equation -x=(y+3)^2+c. Putting x=0.5, y=-3 into the equation, we get -0.5=(-3+3)^2+c, viz., c=-0.5. With experience, you would be able to do this without going through this procedure.
The final equation for the required )-shape parabola is -x=(y+3)^2-0.5. The turning point is x=-0.5,y=-3.

Example 5: [Scaling] To change the shape of the parabola, for instance, to make it "fatter", we write y=k*x^2, where k<1. To make it "thinner", the value of k>1. e.g., both y=x^2 and y=2*x^2 pass through x=0,y=0, but the second parabola is much steeper (or thinner than the first) -- it rises up much more rapidly.

Final comment: In general, a parabola may be written in one of two ways. Let's consider a "U" shape or vertically inverted ("^" shape) parabola. We can write

a) (y-p)=k(x-q)^2+s, OR
b) y=ax^2+bx+c, where p,q,s,a,b,c are all constants.

Given two points on the parabola in a graph, we can uniquely determine an equation which describes the parabola.

e.g., If we know that it passes through (0,1) and (1,4) and it is U-shape (symmetrical about the y-axis), then using formula b, the following conditions must be satisfied:
1=a*(0)^2+b*0+c ...[1] and
4=a*(1)^2+b*1+c ...[2]

From [1], the amount of translation is given by c=1.
From [2], a+b=3. But since the parabola is a mirror image about the y-axis, x=-1,y=4 is also on the curve.

i.e., 4=a*(-1)^2+b*(-1)+c ...[3]
this yields 3=a-b.

Focus on "a+b=3" and "a-b=3". Adding these expressions (side-by-side), (a+b)+(a-b)=3+3 gives 2a=6, a=3. This forces b=0. Thus, the equation is given by y=3x^2+1.

We can readily check that (x=0,y=1) and (x=1,y=4) both satisfy the equation.

Formula a is useful when we know certain properties about the parabola, such as axis of symmetry and turning point, either described in the question or observed from a graph. We can deduce how far a parabola is shifted about the horizontal (x) and vertical (y) axis. This would solve the parameters "p", "q" and/or "s" in (y-p)=k(x-q)^2+s.

Example: If a "^" shape parabola is symmetrical about x=2 and and passes through (x=1,y=5), find its equation in the form (y-p)=k(x-q)^2+s. What is the maximum y value?

Solution: The value of "k" is usually deduced from information regarding the gradient of the parabola at some point. Since you may not have learned about differentiation yet, let us assume that k=-1. This fixes the shape of the parabola. Also, q=2 as x=2 is the axis of symmetry. For this type of parabola, p=0.
Now, if (1,5) lies on the parabola, it must satisfy the equation y=-(x-2)^2+s. This means 5=-(1-2)^2+s .....[1] must be true. Further, since it is symmetrical about x=2, (2,yo) must also satisfy the equation, i.e., yo=s ...[2].

Collectively,
From [1], 5=-1+s ...[3]
From [2], yo=s ...[4]
[4]-[3] gives (yo-5)=1, i.e., yo=6; s=6.
Thus, the maximum point is (x=2,y=6).
The equation is y=-(x-2)^2+6

To understand this:
1) Begin with y=x^2, a "U" shape parabola.
2) Turn this up-side-down, y=-(x^2) [Observe that maximum point is located at (x=2,y=0)]
3) Shift to the right by two units, y=-(x-2)^2.
4) Add a vertical offset to lift the parabola, y=-(x-2)^2+6. Observe that the maximum point is now (x=2,y=6).