Grade 12 Find a Polynomial Given its Graph
Questions with Detailed Solutions
How to find a polynomial given its graph. Grade 12 math questions are presented along with their detailed solutions.

Find the equation of the cubic polynomial function g shown below.
.
Solution
The graph of the function has one zero of multiplicity 1 at x = 1 which corresponds to the factor x + 1 and and a zero of multiplicity 2 at x = 3 (graph touches but do not cut the x axis) which corresponds to the factor (x  3)^{2}, hence function g has the equation:
g(x) = k (x + 1)(x  3)^{2} , where k is a constant.
Constant k may be found using the point with coordinates (1 , 3) shown in the graph.
g(1) = k (1 + 1)(1  3)^{2} = 3
Simplify and solve for k.
k = 3 / 8
g(x) is given by.
g(x) = (3 / 8)(x + 1)(x  3)^{2}

Find the fourthdegree polynomial function f whose graph is shown in the figure below.
.
Solution
The graph of the polynomial has a zero of multiplicity 1 at x = 2 which corresponds to the factor (x  2), another zero of multiplicity 1 at x = 2 which corresponds to the factor (x + 2), and a zero of multiplicity 2 at x = 1 (graph touches but do not cut the x axis) which corresponds to the factor (x + 1)^{2}, hence the polynomial f has the equation:
f(x) = k (x  2)(x + 2)(x + 1)^{2} , where k is a constant.
Constant k may be found using the y intercept f(0) =  1 shown in the graph.
f(0) = k(0  2)(0 + 2)(0 + 1)^{2} = 1
Simplify and solve for k.
k = 1 / 4
f(x) is given by.
f(x) = (1/4)(x  2)(x + 2)(x + 1)^{2}

Find the equation of the degree 4 polynomial f graphed below.
.
Solution
The graph has x intercepts at x = 0 and x = 5 / 2. These x intercepts are the zeros of polynomial f(x). Because the graph crosses the x axis at x = 0 and x = 5 / 2, both zero have an odd multiplicity. The graph at x = 0 has an 'cubic' shape and therefore the zero at x = 0 has multiplicity of 3. The shape of the graph at x = 1/2 is close to linear hence the zero at x = 5 / 2 has multiplicity equal to 1. Using the zeros at x = 0 and x = 5 / 2, f(x) may be written as
f(x) = k (x  0)^{3} (x  5 / 2) , where k is a constant.
We now use the point (2 , 4) to fink k.
 4 = k(2)^{3} (2  5 / 2) , solve for k: k = 1
The equation of polynomial f(x) is given by.
f(x) = x^{3} (x  5 / 2)

The graph of a cubic polynomial $$ y = a x^3 + b x^2 +c x + d $$ is shown below. Find the coefficients a, b, c and d.
.
Solution
The polynomial has degree 3. The graph of the polynomial has a zero of multiplicity 1 at x = 2 which corresponds to the factor x + 2 and a zero of multiplicity 2 at x = 1 which corresponds to the factor (x  1)^{2}. Hence the polynomial may be written as
y = k(x + 2)(x  1)^{2}
We now need to find k using the y intercept (0 , 1) shown in the graph.
1 = k(0 + 2)(0  1)^{2} = 2 k
Solve for k.
k = 1 / 2
We now expand the polynomial, write it in standard form and identify the coefficient a, b, c and d.
y = (1 / 2)(x + 2)(x  1)^{2} = 0.5 x^{3}  1.5 x + 1
We now compare the expression of the polynomial found above to
y = a x^{3} + b x^{2} +c x + d
and obtain the values of the coefficients
a = 0.5 , b = 0 , c = 1.5 and d = 1

The graph of polynomial $$y=ax^4+bx^3+cx^2+dx+e$$ is shown below. Find the coefficients b, d and e.
.
Solution
The graph of the polynomial is symmetric with respect to the y axis and therefore the polynomial function given above must be an even function. The terms b x^{3} and d x included in the given expression of the polynomial above are not even and therefore their coefficients are equal to 0. Hence
b = 0 , d = 0
and therefore y is given by
y = a x^{4} + c x^{2} + e
Coefficient e is found using the y intercept (0 , 2) from the graph.
2 = a (0)^{4} + c (0)^{2} + e
e = 2


More High School Math (Grades 10, 11 and 12)  Free Questions and Problems With Answers
More Middle School Math (Grades 6, 7, 8, 9)  Free Questions and Problems With Answers
More Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers
Author 
email
Home Page