Question

The graph of a cubic polynomial \[y = a{x^3} + b{x^2} + cx + d\] is shown. Find the coefficients \[a,b,c\] and \[d\].

Answer 1

Hint:In the above given question, we are given the graph of a polynomial function which is written as \[y = a{x^3} + b{x^2} + cx + d\] . We have to find the value of the coefficients \[a,b,c\] and \[d\] of the given polynomial function. In the graph of the given function, we can see three points plotted on it, which are \[\left( { - 2,0} \right)\] , \[\left( {0,1} \right)\] and \[\left( {1,0} \right)\] . In order to approach the solution, we can use three points as the points which can satisfy the given equation of the polynomial function since the three points lie on the graph of the function and hence they must satisfy the given equation of the polynomial function.

Complete step by step answer:
Given that, the graph and of a polynomial function whose equation is written as,
\[ \Rightarrow y = a{x^3} + b{x^2} + cx + d\]
From the graph, we can see three points lying on the graph of the given equation, hence they must satisfy the equation of the polynomial function.
The three points lying on the graph are \[\left( { - 2,0} \right)\] , \[\left( {0,1} \right)\] and \[\left( {1,0} \right)\] .
Hence, substituting these values in the given equation, we have
\[ \Rightarrow 0 = a{\left( { - 2} \right)^3} + b{\left( { - 2} \right)^2} + c\left( { - 2} \right) + d\]
That gives us,
\[ \Rightarrow - 8a + 4b - 2c + d = 0\] ...(1)

Again, we have
\[ \Rightarrow 1 = a{\left( 0 \right)^3} + b{\left( 0 \right)^2} + c\left( 0 \right) + d\]
That gives us,
\[ \Rightarrow d = 1\] ...(2)
Similarly, we get
\[ \Rightarrow 0 = a{\left( 1 \right)^3} + b{\left( 1 \right)^2} + c\left( 1 \right) + d\]
That gives us,
\[ \Rightarrow a + b + c + d = 0\] ...(3)
Also the point \[\left( { - 1,2} \right)\] lies on it, hence
\[ \Rightarrow 2 = a{\left( { - 1} \right)^3} + b{\left( { - 1} \right)^2} + c\left( { - 1} \right) + d\]
That gives us,
\[ \Rightarrow - a + b - c + d = 2\] ...(4)

Now, since \[d = 1\] , hence we have the three equations as,
\[ - 8a + 4b - 2c + d = 0 \\
\Rightarrow a + b + c + d = 0 \\
\Rightarrow - a + b - c + d = 2 \\ \]
We can also write it as,
\[ \Rightarrow - 8a + 4b - 2c = - 1 \\
\Rightarrow a + b + c = - 1 \\
\Rightarrow - a + b - c = 1 \\ \]
\[ \Rightarrow 8a - 4b + 2c = 1 \\
\Rightarrow a + b + c = - 1 \\
\Rightarrow a - b + c = - 1 \\ \]
That gives us,
\[ \Rightarrow 8a - 4b + 2c = 1 \\
\Rightarrow a + b + c = a - b + c \\ \]
Hence, we get
\[8a - 4b + 2c = 1 \\
\Rightarrow b = - b = 0 \]

Now since \[b = 0\] and \[d = 1\] ,
Hence, we have \[a + c = - 1\] or \[a = - 1 - c\]
Now since,
\[ \Rightarrow 8a + 2c = 1\]
Hence putting \[a = - 1 - c\] we get,
\[ \Rightarrow 8\left( { - 1 - c} \right) + 2c = 1\]
\[ \Rightarrow - 8 - 8c + 2c = 1\]
That gives us,
\[ \Rightarrow - 6c = 9\]
Hence,
\[ \Rightarrow c = - \dfrac{3}{2}\]
Therefore,
\[ \Rightarrow a = - 1 + \dfrac{3}{2}\]
Hence,
\[ \therefore a = \dfrac{1}{2}\]

Therefore the coefficients of the polynomial function \[y = a{x^3} + b{x^2} + cx + d\] are \[a = \dfrac{1}{2}\] , \[b = 0\] , \[c = - \dfrac{3}{2}\] and \[d = - 1\].

Note:Now since we have found the coefficients of the cubic polynomial function \[y = a{x^3} + b{x^2} + cx + d\] as \[a = \dfrac{1}{2}\] , \[b = 0\] , \[c = - \dfrac{3}{2}\] and \[d = - 1\] , therefore now we can also rewrite the given polynomial function as,
\[ \Rightarrow y = \dfrac{1}{2}{x^3} + 0 \cdot {x^2} - \dfrac{3}{2}x + 1\]
Or just as,
\[ \Rightarrow y = \dfrac{1}{2}{x^3} - \dfrac{3}{2}x + 1\]
Also, it can also be written is the standard form of \[y = a\left( {x - \alpha } \right){\left( {x - \beta } \right)^2}\] where \[\alpha ,\beta \] are the roots of the given equation, as
\[ \Rightarrow y = \dfrac{1}{2}\left( {x + 2} \right){\left( {x - 1} \right)^2}\]
Here, \[\left( { - 2,0} \right)\] and \[\left( {1,0} \right)\] are the roots of a given polynomial function.