Question

which is the graph of the function f(x) = 2x³ - 7x² + 2x + 3?

Explanation:

Step1: Find roots of the function

To find the roots of \( f(x) = 2x^{3}-7x^{2}+2x + 3 \), we can try rational roots. The possible rational roots are factors of 3 over factors of 2, i.e., \( \pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2} \).

Testing \( x = 1 \): \( f(1)=2 - 7+2 + 3=-0\)? Wait, \( 2(1)^3-7(1)^2 + 2(1)+3=2 - 7 + 2+3 = 0 \). So \( x = 1 \) is a root. Then we can factor \( (x - 1) \) from the polynomial. Using polynomial division or synthetic division:

Using synthetic division with root 1:

Coefficients: 2 | -7 | 2 | 3

Bring down 2. Multiply by 1: 2. Add to -7: -5. Multiply by 1: -5. Add to 2: -3. Multiply by 1: -3. Add to 3: 0. So the quotient is \( 2x^{2}-5x - 3 \).

Now factor \( 2x^{2}-5x - 3 \). We need two numbers \( a,b \) such that \( a\times b=2\times(-3)=-6 \) and \( a + b=-5 \). The numbers are -6 and 1. So \( 2x^{2}-6x+x - 3=2x(x - 3)+1(x - 3)=(2x + 1)(x - 3) \). So the roots of \( f(x) \) are \( x = 1,x = 3,x=-\frac{1}{2} \).

Step2: Analyze the end - behavior

For a cubic function \( f(x)=ax^{3}+bx^{2}+cx + d \), when \( a>0 \) (here \( a = 2>0 \)), as \( x
ightarrow+\infty \), \( f(x)
ightarrow+\infty \) and as \( x
ightarrow-\infty \), \( f(x)
ightarrow-\infty \).

Step3: Analyze the graph based on roots and end - behavior

The graph of the function \( f(x)=2x^{3}-7x^{2}+2x + 3 \) should cross the x - axis at \( x=-\frac{1}{2},x = 1,x = 3 \). And since the leading coefficient is positive, the left - end of the graph goes down (as \( x
ightarrow-\infty \)) and the right - end goes up (as \( x
ightarrow+\infty \)).

Answer:

The graph of the function \( f(x)=2x^{3}-7x^{2}+2x + 3 \) is the one that crosses the x - axis at \( x =-\frac{1}{2}\), \( x = 1 \) and \( x = 3 \), with the left - end going down and the right - end going up. (If there are multiple graphs to choose from, the correct one is the one that satisfies these conditions regarding the x - intercepts and end - behavior.)