Finding the Rule

Finding the Rule of Correspondence: a general approach

In many questions on functions, we're given the rule of correspondence for the particular function in question -- however, there are times when we must find the rule either from information in the problem, data in a table or from a graph of the function.

Since each category or family of functions has a "template" format -- we need only plug the values that are given into the parameters or variables they represent, then we can solve for the missing information to complete the definition of the rule of correspondence. Like this:

Example 1:

Find the rule of correspondence for a quadratic function with vertex at ( 1, 3)
through the point P( – 5, – 7).

Solution:

Since we know it is a quadratic function, we'll use the template for quadratics.

f(x) = a(x – h) 2 + k , is the format.

We know h and k since they are the coordinates of the vertex.

So we can rewrite the template replacing h and k with -1 and 3:

f(x) = a(x – (–1)) 2 + 3 = a(x + 1) 2 + 3

Now we only need to find "a".

So, we plug in the x and y values from P(–5, – 7)

since, if P is on the curve, its coordinates must satisfy the rule,

so f(x) = a(x + 1) 2 + 3 becomes – 7 = a(–5 + 1) 2 + 3

since at P(–5, – 7), x = 5 and y or f(x) = 7

Solving for "a", we get So the function rule is .

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In certain questions, the coordinates of a number of points on the curve are presented in table format. Once we know what kind of function is involved, we can define the rule using substitution.

Example 2:

The table below lists the coordinates of points on an Absolute Value Curve.
Determine the rule of correspondence for this function.

 x f(x) 4 7 10 7 7 -3

Solution:

Since we know it is an absolute value function, we'll use the template for that family.

f(x) = a | x – h | + k , is the format.

When we inspect the coordinates of the points in the table, we can see that (4, 7) and (10, 7) are partner points -- they have the same y-value. This tells us that the x-value of the axis of symmetry must be half-way between these x-values since the curve is symmetric to the axis.

The x-value half way between 4 and 10 is 7.

So, the 3rd point in the table (7, – 3) is the vertex (h, k)

since it lies on the axis of symmetry.

Substituting h and k into the template gives:

f(x) = a | x – 7 | – 3

To solve for "a", let's plug-in the coordintes of (4, 7)

7 = a | 4 – 7 | – 3 [ 10 = 3a so  .

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In many word problems, we're presented with a graph of the situation and in solving the problem we must find the rule of correspondence for the curves in the diagram. Again, we read the information that's given, substitute values into the template for the function rule, then solve for the missing pieces.

Example 3:

Depicted below is a Square Root Function. Find the rule of correspondence. The vertex is at (2, 1), the curve moves up and left from the vertex, through P(1, 5).

Since the template for this type of function is: and we know that b = 1 since the curve goes left, our template becomes Now substitute the x and y values from P to find "a" so the function rule is .

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 Function Type Rule Template Quadratic f(x) = a( x – h ) 2 + k, (standard form)f(x) = ax2 + bx + c, (general form) f(x) = a(x – x1)( x – x2), (zeros form) Absolute Value f(x) = a | x – h | + k Square Root Greatest Integer f(x) = a[ b (x – h) ]+ k First Degree Rational f(x) = by division Exponential f(x) = ac ( x - h) + k Logarithmic f(x) = a log c b ( x – h ) + k Transformed Sine f(x) = a sin b(x – h) + k Transformed Cosine f(x) = a cos b(x – h) + k Transformed Tangent f(x) = a tan b(x - h) + k

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Note: there is no mention of the Linear Function in this table since the template is not the most efficient way to find the Linear function rule. The point/slope method discussed in Lesson 3.1 in the Analytic Geometry MathRoom is the only way to find the equation of a line!! In my opinion it is a waste of time and space to go looking for "b" when you already have it. So, once we have , we simply do some algebra to put the equation or function rule in whatever form we need for the question we're doing.

If we want the General Form, we'll cross multiply and set Ax + By + C = 0.

For the Standard Form, we'll multiply through by (x x1) and transpose the y1,
to get y = m (x x1) + y1 .

Once we remove brackets and collect terms, we'll have y = mx + b

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Each of the above functions is discussed fully in a separate lesson in MathRoom Functions.

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1) Find the rule of correspondence for a quadratic function (parabola):

a) with vertex at (1, 1) and y-intercept of 5.

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b) with zeros at 1 and 3, passing through P(7, 16).

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2) Find the function rule for the absolute value function through (2, 5), (0, 5) and (1, 7).

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3) The cross section of a riverbed is a parabola, 60 meters wide with a maximum depth of
6 meters. At what distance from the two shores must we place 2 buoys to mark where the depth is 4 meters? (Hint: use the y-axis as the axis of symmetry.)

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1) Find the rule of correspondence for a quadratic function (parabola):

a) with vertex at (1, 1) and y-intercept of 5.

f(x) = a( x – 1 ) 2 + 1

5 = a( 0 – 1 ) 2 + 1 u a = 4

f(x) = 4( x – 1 ) 2 + 1

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b) with zeros at 1 and 3, passing through P(7, 16).

Since the zeros are 1 and 3, we'll use the zeros form of the function rule.

f(x) = a(xx1)( xx2) u f(x) = a(x + 1)(x – 3)

Now we'll substitute the coordinates of P to solve for "a"

-16 = a(7 + 1)(7 – 3) u a = ½

so f(x) = – ½ (x + 1) (x - 3) in zeros format.

If we want another format, we simply do the algebra.

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2) Note that (2, 5) and (0, 5) have the same y value. Therefore, due to the symmetry of the graph, we know h, the x-coordinate of the vertex is midway between the x-values 2 and 0.
Thus, h = 1, so now we only have to find k and a to complete the question.

We set up 2 equations in 2 unknowns (a and k) and solve.

Using the points (0, 5) and (1, 7) we get:

5 = a | 0 1 | + k [ 5 = a + k
7 = a | 1 1 | + k [ 7 = 2a + k
subtracting, we get 2 = a and substituting a = 2 into 5 = a + k we get k = 3
So the function rule is f(x) = 2 | x - 1 | + 3

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3) A diagram will help immensely here. .

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