We are given four questions and asked to find one that is different from the other three.
Let's highlight the different one based on the answers to the questions.
Find the y-intercept of the graph of y=x^3-2x^2-x+2.
Find the x-intercepts of the graph of y=x^3-2x^2-x+2.
Find all real solutions of x^3-2x^2-x+2=0.
Find the real zeros of f(x)=x^3-2x^2-x+2.
The highlighted question asks us to find a value of y when x=0.
All other questions asks us to set x^3-2x^2-x+2 to 0 and solve the equation.
The three questions ask this using different wording, but they all ask about the same solution set.
We are also asked to find the answer to these questions.
Finding the y-intercept
To find the y-intercept, we substitute x=0 in the equation.
The y-intercept of the graph of y=x^3-2x^2-x+2 is (0,2).
Finding the x-intercepts
To find the x-intercepts, we substitute y=0 and solve the resulting equation.
x^3-2x^2-x+2=0
Since this is a polynomial equation with integer coefficients, we can apply the Rational Root Theorem.
The main coefficient is 1, so the rational roots are all integer factors of the constant term 2.
Let's list all the factors.
± 1, ± 2
There are four possible integer solutions. Let's try all of them.
x
x^3-2x^2-x+2
Value
1
1^3-2( 1^2)- 1+2
0 âś“
-1
( -1)^3-2( -1)^2-( -1)+2
0 âś“
2
2^3-2( 2^2)- 2+2
0 âś“
-2
( -2)^3-2( -2)^2-( -2)+2
12
We found three solutions to the equation.
Since this is a polynomial equation of degree 3, we know that there are at most three real solutions.
This means that we found all three x-intercepts of the graph of y=x^3-2x^2-x+2.
(-1,0), (1,0), (2,0)
