Question

The given graph can be represented by which of the following functions

• A. $y=x^3-3x^2-2x+1$
• B. $y=x^3-2x^2-3x$
• C. $y=x^3+3x^2-2x+1$
• D. $y=x^3-2x^2+3x$

Answer 1

The correct option is A $y=x^3-3x^2-2x+1$

By seeing the graph we can say that $y$-intercept at $y\left(0\right)=1$

1) First of all, we need to find the $x$-intercept of the polynomial function.

$y\left(0\right)=0^3-3\times 0^2-2\times 0+1=1$

$y\left(0\right)=0^3+3\times x^0-2\times 0+1=1$

Options B and D do not have y-intercept at $y=1$.

3) If function is a polynomial function of degree $n$, then there is at most $n-1$ turning point on the graph of function.

Degree of the polynomial function will be $3$

So, there will be at the most $3-1=2$ turning point of the given graph.

4) Determine the symmetry

$y$-axis symmetry:

$f\left(x\right)=x^3-3x^2-2x+1$

Substitute the value of $x=-x$

$f(-x)=-x^3-3x^2+2x+1=-(x^3+3x^2-2x-1)$

So, $f(-x)\ne f\left(x\right)$

Symmetric about the origin:

$f(-x)=-f\left(x\right)$

It does not statisfy the condition.

5) Here, leading coefficient is $1$ which is odd and degree of the polynomial function will be $3$ which is odd.

Therefore, we can say that graph of the function fall to the left and rises to the right.

6) Find some extra points which are required in that question

$f\left(1\right)=1^3-3\times 1^2-2\times 1+1=-3$(Option A)

$f\left(1\right)=1^3+3\times 1^2-2\times 1+1=3$(Option C)

So, option A satisfies the condition $f\left(1\right)=-3$.